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NEW HIGH SCHOOL 
ARITHMETIC 


REVISED 


BY 

WEBSTER WELLS 

u 

AND 

WALTER W. HART 

t 

ASSOCIATE PROFESSOR OF MATHEMATICS 
SCHOOL OF EDUCATION 
AND 

TEACHER OF MATHEMATICS 
WISCONSIN HIGH SCHOOL 
UNIVERSITY OF WISCONSIN 



? * 

i i» 


D. C. 

HEATH AND 

COMPANY 

BOSTON 

NEW YORK 

CHICAGO 

ATLANTA 

SAN FRANCISCO 

DALLAS 


LONDON 



o 










Copyright, 1933, 

By Emily R. Wells and Walter W. Hart 


No part of the material covered by this 
copyright may be reproduced in any form 
without written permission of the publisher. 


This book may be had 
with answers or without answers 
at the same price. 

Answer books, bound in paper, may be 
obtained free of charge by teachers. 

3 A 3 


Printed in the United States of America 


MAR 10 i333 

©CIA 59900 




PREFACE 


The high school course in arithmetic is unique in its educa¬ 
tional possibilities. It is rich in its results in general mind and 
character training. For all pupils who take it, it develops skill in 
computation, and furnishes indispensable knowledge of business 
and industrial processes and facts that are not acquired in the 
lower grades or elsewhere in school. For many pupils, such skill 
and knowledge are distinctly necessary preparation for vocations 
which they enter later; the girl who becomes a teacher, the boy 
who becomes a farmer or mechanic, the boy or girl who enters any 
form of commercial work — these all find arithmetical skill and 
the information acquired in solving the problems invaluable. 

The New High School Arithmetic furnishes material for such 
instruction. It is adapted to the needs of all the pupils, whether 
they take a general course, a normal course, an industrial voca¬ 
tional course, or a commercial course. 

It is modern and varied in its appeal because of the use of a 
great variety of problem material drawn from current life. Ob¬ 
serve the variety of material used in the problems on pages 10-11, 
19-22, 41-43, 212-214, 216-220, 230-233; the abundance of prob¬ 
lems having vocational interest for boys, such as 43^8 on pages 61- 
62, 46-47 on page 65, 53-55 on pages 73-74, 6-13 on page 105, all of 
Chapter VI, 33 on page 231, etc.; the domestic science problems 
such as 47-50 on page 70, 51-52 on page 73, 23-25 on page 77, etc.; 
the appeal to the interest of the country pupils as well as those in 
the city as in example 19 on page 76, 22 on page 77, 25-30 on 
page 121, etc.; the effort to arouse interest in national and local 
affairs as in examples 31-36 on pages 98-99, § 144 on page 201, 
§ 196 on page 292, etc. 

It is accurate and entirely adequate in its teaching of business 
and industrial practices. Note, as examples, the tabulations of 
statistical material (pp. 14-16) ; bills on pages 34-37; estimating 

iii 




IV 


PREFACE 


costs on pages 153, 154, 156, 179-188; and the full treatment of 
“ commercial arithmetic ” in Chapters VIII to XIII inclusive. 

In its pedagogy this text is like the other texts of the Wells and 
Hart series. It is mathematically accurate; note particularly 
§ 98 on page 145, § 100 on page 147, and § 113 on page 165. The 
thought processes involved in solutions are developed clearly in an 
inductive manner; see § 49 on page 67, § 50 on page 71, § 107 on 
page 158, etc. The processes are clearly and strikingly expressed 
in unique form in rules such as those on pages 30, 138, 255, etc. 
The model solutions set an example of clearness and order that 
cannot fail to impress pupils and result in more accurate computa¬ 
tion by them ; see pages 44, 45, 75, 108, 269, 308, 311, etc. New 
important terms are printed in striking manner (page 17) and are 
fully indexed. There is provided for drill purposes an abundance 
of examples and problems (pp. 5, 6, 8, 39, 58, 235) which are of 
the simple sensible sort met in real applications (pp. 55, 56, 57, 
59, 63,103). These examples can and should often be done men¬ 
tally, and will be found to be properly graded in difficulty in the 
separate lists. Topics which are not strictly essential for all pupils 
are plainly marked as supplementary. These topics are in some 
cases interesting but not necessary (as Apothecaries’ Measure on 
page 132), and in other cases are important but can be included in 
a course only when local conditions render it possible (as Topic A 
on page 179). 

How much and what to include in the course in arithmetic in 
a particular community depends upon the previous preparation 
of the pupils, upon the special aims of the course, and upon the 
time allowed for the study. In most cases some selection of topics 
as well as of examples and problems becomes necessary. 

Chapters I to V, excepting the Supplementary Topics, are a 
fundamental part of any course. An effort should be made, how¬ 
ever, to get over these chapters as speedily as is consistent with 
the development of a reasonable amount of skill in computation. 
An abundance of drill has been provided, with the expectation 
that some of it will be done orally and some of it will be used for 
special assignments for those pupils who are weak in computation. 
Obviously it is not necessary that all the examples and problems 
should be solved by any one class. 

What use is made of Chapter VT will depend upon whether the 


PREFACE 


V 


course in arithmetic precedes or follows the course in geometry. 
If it follows, then much of this chapter between pages 145 and 
168 should be omitted, although some of the problems should be 
solved. If time must be saved, pages 168 to 197, and 205 to 207 
may also be omitted. If this chapter precedes the course in 
geometry, it will be found valuable preparation for that course. 






CONTENTS 


1. 

Integers . 

PAGE 

1 

II. 

Fractions. 

. 51 

III. 

Decimal Fractions. 

. 85 

IV. 

Denominate Numbers .... 

. 108 


Supplementary Topics .... 

. 123 


A. The Metric System of Measures . 

. 123 


B. Foreign Money ..... 

. 130 


C. Apothecaries’ Measures.... 

. 132 

V. 

Involution and Evolution 

. 134 


Supplementary Topic — Using Powers and Roots 


in Formulas. 

. 142 

VI. 

Practical Geometry and Applications . 

. 145 


Supplementary Topics .... 

. 179 


A. Problems of the Carpentering Contractor 

. 179 


B. Mensuration of Less Common Solids 

. 189 


C. Graphical Representation 

. 197 


D. United States Government Land Survey 

. 205 

VII. 

Percentage . 

. 208 

VIII. 

Buying and Selling Goods 

. 226 


Successive Discounts .... 

. 227 


Profits and Losses. 

. 234 


Marking Goods. 

. 237 


Commission and Brokerage . 

. 239 

IX. 

Interest . 

. 245 


Compound Interest. 

. 258 


Supplementary Topics .... 

. 262 


Postal Savings Banks .... 

. 262 


vii 










CONTENTS 


Vlll 


X. 

XI. 


XII. 


XIII. 


Compound Interest Table 



PAGE 

. 265 

Periodic Interest .... 



. 267 

Installment Payment of Debts . 



. 269 

Insurance. 



. 272 

Taxes. 



. 287 

Supplementary Topics 



. 292 

State Income Taxes 



. 292 

State Inheritance Taxes 



. 294 

Duties and Customs 



. 295 

Internal Revenue .... 



. 298 

United States Income Taxes . 



. 298 

United States Inheritance Taxes 



. 301 

Banking. 



. 303 

Promissory Notes .... 



. 303 

Partial Payments .... 



. 307 

Discounting Notes .... 



. 313 

Exchange — Checks .... 



. 317 

Drafts and Acceptances 



. 322 

Foreign Exchange .... 



. 331 

Partnerships and Corporations 



. 335 

Stocks and Bonds .... 



. 339 

Index. 













NEW 

HIGH SCHOOL ARITHMETIC 


I. INTEGERS 

1. Writing and Reading Integers in the Hindu System.— 

Pupils are familiar with the manner of writing and reading 
small Integers or whole numbers in the Hindu System by 
means of the digits 0, 1, 2, 3, •••, 9. 

Note. — Our system of writing numbers is often improperly 
called the Arabic System. 

They should recall that a number like 2,643 is read “ two 
thousand, six hundred forty-three” not “ and forty-three.” 

EXERCISE 1 

1. Read 4,275; 3,560; 2,708; 6,029. 

2. What are the digits ? 

3 . Tell what each digit of the number 6,035 represents. 

4 . In numbers like fourteen, what does the syllable 
“teen” mean? 

5 . In numbers like thirty, what does the syllable “ty” 
mean ? 

2. For convenience in reading integers, the digits of a 
number are divided into groups of three, starting from the 
right. Each group is called a Period. The first three figures 
form the units’ period; the second three, the thousands 

l 


2 


NEW HIGH SCHOOL ARITHMETIC 


period; the third three, the millions’ period; the fourth 
three, the billions’ period. 

Thus, 20,793,106,452 is read twenty billion , seven hundred ninety - 
three million , one hundred six thousand , four hundred fifty - two . 

EXERCISE 2 

Read the statements in the following examples. 

1. a. Maine is the largest of the New England States, 
having an area of 33,040 square miles. 

b. Pennsylvania is the largest of the Middle Atlantic 
States. Its area is 96,699 square miles. 

c. Georgia is the largest of the South Atlantic States, 
with an area of 59,265 square miles. 

2. The total population of the United States according to : 

a. The census of 1890 was 62,947,714. 

b. The census of 1900 was 75,994,575. 

c. The census of 1910 was 91,972,266. 

d. The census of 1920 was 105,710,620. 

e. The census of 1930 was 122,775,046. 

3 . The state of the United States having the largest area 
is Texas. It covers 265,896 square miles. 

4 . The city of the United States having the largest popu¬ 
lation in 1930 was New York. At that time its population 
was 6,930,446. 

5. The population of London in 1931 was 8,202,818. 

6. a. In the year 1928 there were 1071 universities, col¬ 
leges, and other schools for higher education in the United 
States. 

b. In these schools for higher education there were 
402,242 men students and 292,977 women students. 

7 . During the same year there were 2,576,157 pupils in 
private and parochial schools in the United States. 

8. In 1930 there were 4,283,753 persons in the United 
States, 10 years of age and older, who were illiterate. 


INTEGERS 


3 


9 . The states of the United States leading in the produc¬ 
tion of certain crops in 1930 were: 

a. Texas, which produced 4,100,000 bales of cotton, 
valued at $192,700,000. 

b. Maine, which produced 45,120,000 bushels of potatoes, 
valued at $ 29,328,000. 

c. North Carolina, which produced 535,195,000 pounds of 
tobacco, valued at $ 73,322,000. 

d. Iowa, which produced 360,750,000 bushels of corn, 
valued at $ 209,235,000. 

e. Georgia, which produced 384,200,000 pounds of pea¬ 
nuts, valued at $12,679,000. 

10. a. In 1928 there were 30,887,167 children in the 
United States. 

6. Of these, 25,179,696 were enrolled in public schools. 

c. 831,934 teachers were employed to teach them. 

d. The total expense of the public schools was 
$2,180,558,660. 

3. The Roman System of writing integers. 

I represents 1 C represents 100 

Y represents 5 D represents 500 

X represents 10 M represents 1000 

L represents 50 

All other integers are represented by writing two or 
more of these letters side by side. 

Rule I. — If a letter follows one of the same or greater 
value, the number represented is the sum of their values. 

Thus, LXVII = 50 + 10 + 5 + 2, or 67. 

Rule II. — If a letter precedes one of greater value, the 
number represented is the difference between their values. 

Thus, XC = 100 - 10, or 90. 

Rule III. —Always use Rule II first, and afterward Rule I. 

Thus, to read MCMXL, first read CM as 900, and XL as 40 ; 
then M (CM) (XL) is 1000 + 900 + 40, or 1940. 


4 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 3 

In Examples 1 to 14, read the numbers and dates. 

1. YII 3 . IX 5 . XVIII 7 . XXVII 9 . DCXI 

2 . XL 4 . CM 6 . MDC 8 . LXIX 10 . MDXC 

11. Columbus discovered America in MCDXCII. 

12. Washington was born in MDCCXXXII. 

13 . Lee was born in MDCCCVII. 

14 . The World War started in MCMXIV and ended in 
MCMXVIII. 

Addition of Integers 

4. The, integers arranged in the order 1, 2, 3, 4, etc., are 
said to form a Sequence. 

The Sum of two integers like 3 and 5 is the fifth integer 
beyond 3 in the sequence of integers. 

Addition of Integers is the process of finding the sum when 
two or more integers are given. The integers added are 
called Addends. 

5. Skill in Adding depends upon accurate and rapid recog¬ 
nition of the sums of the following pairs of integers. 

1 051437214 

L AAJLJLA1_JL JLJ 

3615685962 

2 191212347 

?__L_2AA_^!_AAJL 

9 823235382 

L_2JLA_1AJL-L_L3 

6 711362478 

5 494635223 







INTEGERS 


5 


5 

4 

4 

8 

3 

6 

5 

9 

4 

4 

2 

8 

3 

4 

8 

4 

3 

5 

7 

5 

7 

1 

5 

0 

9 

0 

0 

8 

0 

6 

3 

0 

9 

6 

8 

3 

5 

6 

9 

_7 

8 

9 

2 

0 

3 

6 

0 

5 

0 

4 

7 

0 

0 

7 

9 

9 

4 

7 

2 

9 

0 

7 

4 

8 

9 

0 

7 

6 

9 

5 

8 

5 

0 

5 

6 

1 

9 

8 

7 

8 

7 

5 

9 

3 

7 

8 

9 

6 

8 

7 

8 

0 

3 

0 

6 

0 

4 

0 

9 

0 


EXERCISE 4 


Addition Drill Table 

Note. — The following table can and should be used for a 
variety of drill. The following exercises are suggestive. It is 
not necessary that all be done at this time or that every pupil do 
all of them. On the other hand, any pupil who is inaccurate or 
slow in adding should be required to practice on this drill daily 
until he can do the exercises skillfully. From time to time, a few 
minutes can be spent profitably upon oral drill based on this table. 

Do not copy the numbers from the book. 

Give the results orally or write only the results, when possible. 



a 

b 

c 

d 

e 

/ 

9 

h 

i 

a. 

16 

56 

86 

46 

76 

66 

26 

36 

96 

b. 

17 

27 

47 

37 

67 

57 

97 

87 

77 

c. 

18 

38 

48 

28 

58 

78 

68 

88 

98 

d. 

29 

19 

49 

59 

39 

69 

89 

99 

79 

e. 

38 

17 

46 

53 

75 

64 

85 

92 

24 

/• 

57 

45 

26 

19 

74 

33 

68 

87 

98 

9- 

16 

37 

58 

27 

49 

67 

76 

95 

86 

h. 

47 

66 

85 

54 

39 

77 

96 

23 

18 

i. 

78 

93 

48 

64 

36 

24 

59 

15 

88 

j • 

56 

13 

79 

28 

94 

65 

89 

32 

43 






6 


NEW HIGH SCHOOL ARITHMETIC 


1. To each number of row a, add 

a. 4 b. 3 c. 5 d. 9 e. 7 /. 6 8 

2 - 10 . Add these same numbers successively in rows 
b, c, d, - j . 

11 - 19 . Add these same numbers successively to the num¬ 
bers in columns a, b, c, ••• i. 

20 - 29 . Add 11 to the numbers in rows a, b, c, ••• J. 

30 - 39 . Add 12 to the numbers in rows a, 6, c, ••• j. 

6. Some Devices Employed in Adding. 

a. When adding columns, name results only. 

6 

j Think: 9, 16, 22. Write 22. 
g Do not think: 4 and 5 are 9 ; 

9 and 7 are 16 ; 

— 16 and 6 are 22. 

22 


EXERCISE 5 
Addition Drill Table 


Note. — Read the note at the beginning of Exercise 4. 


abed 

a. 5 7 8 6 

b. 3 4 5 7 

c. 8 3 9 2 

d. 4 5 6 8 

e. 9 6 4 3 

/. 2 8 3 9 

g. 6 9 2 5 

fc. 7 2 7 4 

f. 8 7 5 6 

j. 3 6 8 7 


e f q hi 

2 3 4 9 7 

6 5 8 3 2 

5 4 3 7 6 

7 6 5 8 4 

4 7 6 5 8 

8 9 7 6 3 

3 8 2 4 5 

9 2 9 2 9 

4 7 8 5 8 

6 5 4 6 5 


j 

6 

9 

5 

3 

7 

8 
2 

4 

6 

5 


INTEGERS 


7 


1. Hold a slip of paper below row b. Give orally or 
write upon the paper the sum of the two numbers in each 
column that appear above the paper. 

2 - 9 . Similarly, give orally or write upon slips of paper 
the sum of the three numbers in each column that are in 
rows, a, b, and c ; of the four numbers in each column that 
are in rows a, b , c, and d j etc. 

10 - 18 . Similarly, find the sums of the numbers in each 
column that are in the lower two rows, the lower three 
rows, etc. 

Note. — By omitting row a, and following the directions for 
Examples 1-9, other examples can be made; also, by omitting 
row j, and following the directions for Examples 10-18, more 
examples can be provided; or, by adding a row at the top of the 
table, new examples are provided. 

b. When adding columns of integers, group and add at sight 
integers whose sum is 10. 

These numbers may not be consecutive in the column. 



'27 Think: 10,18,25,35. Write 5 and carry 3, 
^98 Think: 7, 17, 27, 37. Write 37. 


375 


In the same manner, group and add at sight integers 
whose sum is readily recognized. Thus, if 5 and 3 appear 
in sequence in a column, instead of adding them separately, 
add 8 at once. Skill in making such combinations comes 
with practice and greatly increases the student’s speed in 
adding. 



8 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 6 

Add the following columns, grouping numbers whose sum 
is 10: 


b 
1 
9 
6 

4 

5 
5 
3 
2 
8 

8 5 
6 2 
4 8 


h i 


8 8 


l m n 


EXERCISE 7 

Addition Drill 


Read the note at the beginning of Exercise 4. 




a 

b 

c 

d 

e 

/ 

9 

h 

i 

j 

a. 

223 

746 

637 

659 

63 

477 

785 

693 

762 

781 

b. 

452 

863 

708 

794 

872 

536 

439 

265 

574 

626 

c. 

537 

227 

430 

664 

580 

229 

546 

537 

435 

543 

d. 

365 

308 

546 

573 

85 

306 

757 

729 

918 

567 

e. 

248 

549 

723 

685 

37 

540 

368 

456 

832 

112 

f 

374 

802 

862 

746 

709 

907 

472 

862 

767 

346 

9 • 

889 

980 

995 

334 

430 

823 

23 

578 

648 

752 

h. 

496 

679 

726 

628 

667 

321 

94 

384 

953 

374 

i. 

534 

863 

849 

792 

82 

530 

506 

724 

165 

439 

j- 

263 

421 

741 

873 

280 

865 

729 

932 

718 

326 


1-20. See the examples suggested for Exercise 5, page 7. 
21 . Find the sum in each column of the numbers in rows 
b, c, d, and e ; in rows c, d, e, f, and g ; etc. 


INTEGERS 


9 


c. When long columns are to be added, the units’ column 
may be added and the sum written ; then the tens’ column may 
be added and the result written as in the following example; 

etc. 

67 

The sum of the units = 30 units = 30 

o . q The sum of the tens = 21 tens = 210 

The sum of the hundreds = 22 hundreds = 2200 
The complete sum = 2440 

Note. — If one who is adding in this manner is interrupted, he 
need not repeat the whole addition. 

EXERCISE 8 


Add upward the following columns, and verify your 
results by adding again downward: 


1 . 

2 . 

3 . 

4 . 

5 . 

1789 

5403 

4529 

7854 

1827 

6543 

786 

7992 

6215 

4329 

2177 

9230 

467 

9448 

25070 

915 

1157 

8920 

4007 

6118 

6783 

8980 

3508 

651 

2522 

325 

7526 

2463 

5869 

3909 

6 . 

7 . 

8 . 

9 . 

10 . 

79856 

45340 

43765 

58964 

77167 

35117 

10087 

89140 

86723 

63489 

32949 

76322 

35174 

48213 

68791 

18817 

36450 

64385 

54876 

74153 

85622 

78809 

79160 

83538 

84375 

16304 

39713 

23099 

71766 

97561 


11. The prize cow of a certain herd produced milk as 
follows : Monday 37 lb.; Tuesday 36 lb.; Wednesday 40 lb. ; 
Thursday 38 lb.; Friday 37 lb.; Saturday 39 lb.; Sunday 
38 lb. What was the production for the week? 











10 


NEW HIGH SCHOOL ARITHMETIC 


12. A grocer received in cash, on Monday $ 135.25, on 
Tuesday $84.40, on Wednesday $106.65, on Thursday 
$ 122.70, on Friday $ 93.62, and on Saturday $ 185.66. 
What were his total receipts for the week ? 

13 . John’s father asked him to find how much money 
had been taken in at their store on Saturday in checks. 
The amounts on the checks were : $ 17.75, $ 12.43, $22.94, 
$13.56, $18.27, $29.85, $6.70, $13.72, $6.94, $4.83, 
$8.24, $5.48. 

You also find the sum. 

14 . The treasurer of the athletic association had the fol¬ 
lowing record of receipts from five football games. What 
was the total ? 1st game, $ 27.50; 2nd game, $ 32.25; 
3rd game, $ 18.75 ; 4th game, $25.00 ; 5th game, $ 28.75. 

15 . The receipts at the school lunch room one month 
were : 1st week, $ 118.47 ; 2nd week, $ 123.52 ; 3rd week, 
$ 121.83 ; 4th week, $ 126.35. What were the receipts for 
the month ? 

16 . In a “ drive ” to collect money for the Community 
Relief Fund: Team A collected $754.35; Team B, 
$ 1530.65 ; Team C, $ 2475.25 ; Team D, $ 946.15 ; Team E, 
$1848. What was the total collected by these teams ? 

17 . The following bills were paid by a house owner for 
the construction and equipment of his new home: carpen¬ 
ter, $3016.77; mason, $1500; plasterer, $674; tinsmith, 
$94.25; plumber, $825.75; heating contractor, $623.50; 
electrician, $ 174.88 ; hardware merchant, $ 195.00. Tabu¬ 
late and find the total. 

18 . During 1930 the larger railroads of this country 
made the following additions to their equipment: for 
locomotives, $ 88,494,000 ; for freight cars, $ 181,028,000 ; 
for passenger cars, $44,791,000; for other equipment, 
$ 13,956,000. What was the total amount spent for such 
additional equipment in 1930? 


INTEGERS 


11 


19. Arrange in a table and find the totals of the sales and 
expenses (including salary) of each of five agents of a com¬ 
pany. A: expenses $ 6800, sales $ 42,500 ; B : expenses 
$ 6100, sales $ 41,000 ; C : expenses $ 6850, sales $ 47,400 ; 
D : expenses $ 6550, sales S 53,250; E : expenses $ 6325, 
sales $46,850. 

20. The following table gives the land and water areas 
of each of the ten largest states. Find the totals. 


State 

Land 

Water 

Texas. 

262,398 sq. mi. 

3498 sq. mi. 

California .... 

155,652 

2645 

Montana .... 

146,131 

866 

New Mexico . . . 

122,503 

131 

Arizona. 

113,810 

146 

Nevada . 

109,821 

869 

Colorado .... 

103,658 

290 

Wyoming .... 

97,548 

366 

Oregon. 

95,607 

1092 

Utah. 

82,184 

2806 


21. Below are the names of the five states which pro¬ 
duced the largest amounts of cotton in a recent year, the 
number of bales each produced, and the value of the cotton. 
Arrange them in a table, and find the totals. 

S. Carolina: 1,040,000 bales, $51,480,000; Georgia: 
1,625,000 bales, $ 75,562,000; Alabama: 1,495,000 bales, 
$67,275,000; Mississippi: 1,500,000 bales, $75,000,000; 
Texas : 4,100,000 bales, $ 192,700,000. 

22. During the year 1930 to 1931, the income taxes 
collected by the United States in each of certain four 
quarters were: 1st quarter, $553,725,824; 2nd quarter, 
$553,586,266; 3rd quarter, $398,541,064; 4th quarter, 
$ 354,348,484. What was the total amount of income taxes 
collected in the four quarters ? 














12 


NEW HIGH SCHOOL ARITHMETIC 


d. Mental Addition. — The following example illustrates 
a useful form of mental addition. 

Example. Add 34 and 59. 

Solution. Think 34 + 50 = 84; then 84 + 9 = 93. 

EXERCISE 9 
Find the following sums mentally. 


1. 

43 + 62 

7. 39 + 54 

13. 

82 + 33 

2. 

27 + 38 

8. 28 + 69 

14. 

93 + 58 

3. 

36 + 57 

9. 42 + 87 

15. 

44 + 67 

4. 

83 + 55 

10. 57 + 38 

16. 

37 + 67 

5. 

56 + 73 

11. 52 + 66 

17. 

75 + 46 

6. 

68 + 94 

12. 75 + 49 

18. 

89 + 29 


e. Horizontal Addition. — If small numbers to be added 
happen to be arranged horizontally, they should be added 
without taking time to rewrite them in a column. 

Example. Find the following sum : 

55 + 36 -1-24 + 47 4- 63 

Solution. 1. Adding the ones , think: 5, 11, 15, 22, 25. Write 
5 and carry 2. 

2. Adding the tens, think: 7, 10, 12, 16, 22. Write 22 in front 
of the 5. 

3. .*. 55 + 36 + 24 + 47 + 63 = 225. 


EXERCISE 10 


Find the following sums without copying the numbers. 


l. 

26 + 43 + 65 + 34 

8. 

2. 

47 + 32 + 27 + 56 

9. 

3. 

68 + 72 + 85 + 47 

10. 

4. 

32 + 55 + 43 + 67 

11. 

5. 

74 + 97 + 25 + 89 

12. 

6. 

16 + 69 + 57 + 71 

13. 

7. 

38 + 81 + 49 + 94 

14. 


233 + 152 + 415 + 376 
528 + 365 + 147 + 283 
694 + 268 + 153 + 562 
367 + 489 + 532 + 794 
589 + 612 + 755 + 927 
722 + 845 + 988 + 251 
955 + 378 + 123 + 585 


INTEGERS 13 

15. Complete the following record of hours worked by 
the men in one department in a shop. Check. 


Workman 

M 

T 

w 

Th 

F 

s 

Total 

John. 

8 

‘ 8 

8 

9 

8 

4 


Arthur .... 

8 

8 

6 

- 

- 

4 


Charles .... 

4 

8 

8 

8 

4 

4 


William . . . 

- 

4 

8 

8 

8 

4 


Frank .... 

8 

4 

4 

4 

6 

4 


James .... 

8 

9 

9 

8 

9 

4 


Total .... 









16. The manager of a department prepared the following 
report of sales made by his clerks. Complete and check. 


Sales Person 

Cash Sales 

Charge Sales 

Total 

Miss Flynn . . 

$42.56 

$78.45 


Miss Andrews . 

33.85 

43.70 


Mr. Porter . . 

24.70 

55.65 


Miss Benson 

56.50 

75.15 


Mr. Harris . . 

35.48 

55.25 


Total .... 





17. In a campaign to raise funds, five teams obtained the 
following sums in one week. Complete and check. 


Team 

Mon. 

Tues. 

Wed. 

Thurs. 

Fri. 

Sat. 

Total 

A 

$250 

$187 

$175 

$210 . 

$230 

$255 


B 

314 

285 

270 

283 

300 

325 


C 

108 

92 

76 

87 

^101 

98 


D 

524 

465 

423 

386 

' 672 

509 


E 

132 

143 

107 

196 

184 

165 


Total 



















































14 


NEW HIGH SCHOOL ARITHMETIC 


18. Ten lamp cleaners of a gas company cleaned the in¬ 
dicated numbers of lamps on the six days of a week. Find 
the total each day for all the cleaners and the total for the 
week for each cleaner. 


Employees 

Mon. 

Tues. 

Wed. 

Thurs. 

Fri. 

Sat. 

Total 

John Ewing .... 

50 

54 

53 

60 

58 

55 


Alfred Bower . . . 

15 

18 

22 

29 

25 

33 


Harry Smith .... 

32 

37 

41 

30 

39 

43 


James Swenson . . . 

60 

62 

63 

60 

61 

59 


Roy Crane .... 

40 

45 

43 

49 

51 

47 


Tim Sullivan . . . 

57 

52 

46 

40 

39 

50 


Lee Anderson . . . 

50 

47 

41 

45 

49 

48 


Jacob Brown . . . 

80 

85 

70 

82 

75 

50 


Paul Derby .... 

12 

21 

30 

35 

41 

32 


Andrew Ritz .... 

27 

35 

41 

47 

39 

42 


Total . 









19. At a milk condensery, certain farmers on the days 
of one week sold the number of pounds of milk indicated. 
Find the total each day for all the farmers and the total 
for the week for each farmer. 


Farmer 

Mon. 

Tues. 

Wed. 

Thurs. 

Fri. 

Sat. 

Sun. 

Total 

A 

125 

128 

123 

130 

126 

121 

123 


B 

880 

876 

882 

879 

876 

883 

877 


C 

231 

232 

235 

227 

229 

236 

230 


D 

240 

235 

236 

241 

243 

237 

238 


E 

270 

260 

265 

273 

262 

268 

271 


F 

1162 

1158 

1156 

1160 

1163 

1159 

1155 


G 

465 

459 

462 

468 

470 

460 

464 


H 

932 

935 

934 

929 

931 

936 

933 


I 

532 

533 

528 

535 

529 

531 

530 


J 

763 

762 

760 

757 

765 

766 

759 


Total 












































INTEGERS 


15 


20. Copy and complete this pay roll. 


Workman 

M 

T 

w 

Th 

F 

s 

Total 

John . . . 

5 

6 

5 

5 

6 

4 


Sam .... 

5 

8 

5 

4 

5 

4 


Edward . . 

4 

5 

5 

5 

5 

4 


Charles . . 

6 

6 

6 

8 

8 

4 


Andrew . . 

8 

8 

8 

8 

0 

0 


Peter . . . 

0 

0 

4 

5 

5 

4 


Total 









21. Complete the following record of sales. 


Clerk 

Cash Sales 

Charge Sales 

Total 

A 

$ 25.68 

$ 48.24 


B 

73.62 

64.80 


C 

125.43 

175.52 


D 

84.96 

105.43 


E 

90.78 

83.65 


Total . 





22. Complete the following record of several depositors 
in a bank. 


Depositors 

Opening 

Balance 

Deposits 

Total 

Checks 

Drawn 

Closing 

Balance 

A 

$937 

$265 


$415 


B 

672 

138 


95 


C 

756 

325 


284 


D 

342 

98 


235 


E 

166 

550 


430 


F 

826 

155 


366 


G 

598 

231 


187 


H 

462 

113 


286 


Total . 























































16 


NEW HIGH SCHOOL ARITHMETIC 


23. Household expenditures of a family. Find the total 
for each month and the yearly total for each item. 


Months 

Coal 

Life and 
Fire In¬ 
surance 

Taxes 

Doctors 

AND 

Drug¬ 

gists 

House 

Account 

Clubs 

Recrea¬ 

tion 

Total 

Jan. 

$124.50 

$15.05 

$53.95 

$64.42 

$ 85.00 

$ 7.50 

$11 50 


Feb. 


85.63 


4.00 

90.19 

2.50 

7.25 


Mar. 

16.50 


9.50 


97.04 


8.00 


Apr. 


15.05 


5.00 

87.15 


6.00 


May 


27.00 


2.85 

83.17 

15.00 

8.30 


June 




13.75 

98.45 


4.25 


July 


15.05 


7.67 

83.75 


9.00 


Aug. 





118.44 

25.00 

5.35 


Sept. 


39.40 


5.00 

85.77 


55.00 


Oct. 


15.05 



87.93 


5.75 


Nov. 




6.75 

90.25 

2.70 

9.40 


Dec. 




3.63 

100.65 

5.00 

15.00 


Total 










24. The bookkeeper of A. C. Cole prepared the following 
summary of the year’s business. Find the totals. 


Month 

Total Sales of 

Coal 

Wood 

Ice 

Lime 

Cement 

Total 

January. . . 

$12,500 

$245 

$ 250 

$150 

$ 25 


February . . 

14,785 

320 

275 

85 

78 


March . . . 

11,690 

165 

440 

246 

164 


April .... 

10,850 

185 

1062 

580 

275 


May .... 

7,270 

200 

2480 

625 

583 


June .... 

4,340 

115 

5675 

840 

962 


July .... 

7,675 

125 

8342 

972 

995 


August . . . 

8,940 

195 

9730 

863 

1145 


September . . 

10,845 

250 

8650 

985 

1067 


October . . . 

10,420 

275 

3240 

745 

965 


November . . 

13,680 

305 

1035 

585 

1084 


December . . 

17,690 

265 

469 

239 

215 


Total .... 
























































INTEGERS 


17 


Subtraction of Integers 

7. In Subtraction, two numbers are given, the Subtrahend, 
and the Minuend. Subtraction is the process of finding the 
number, called the Difference, which must be added to the 
subtrahend to give the minuend. 


Thus, to subtract 8 from 12 means to find the number which 
must be added to 8 to give 12. This is 4. 

Subtraction is also thought of as a process of taking 
aivay and pupils are doubtless more familiar with it in that 
form. 

In actual business computations, both meanings of sub¬ 
traction find application. The following example illustrates 
a solution in accordance with the first meaning. 


Example 1. Subtract 483 from 758. 


Minuend 758 
Subtrahend 483 
Difference 275 


Think 3 + 5=8. Write 5 in the units’ 
place. 

Think 8 + 7 = 15. Write 7 in the tens’ 
place. 

Think 1, carried, + 4 = 5 + 2 = 7. Write 
2 in the hundreds’ place. 


Check: The sum of the subtrahend and difference equals the 
minuend. 


Example 2. Subtract 436 from 859 by the taking away 
method and explain the solution. 

Whatever the method used, skill in subtraction, like 
skill in addition, depends upon accurate and rapid recogni¬ 
tion of the difference between (a) any integer less than ten 
and a greater integer to ten inclusive, or ( b ) between any 
such integer and the integer formed by adding ten to a 
smaller integer. 


18 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 11 

Find the differences indicated in the last sentence of the 
preceding paragraph. There are eighty-one such differences. 





EXERCISE 

12 




a 

b 

c 

d 

e 

/ 

9 

h 

i 

a. 21 

43 

37 

66 

58 

74 

88 

79 

92 

b. 44 

95 

24 

35 

89 

76 

62 

53 

18 

c. 87 

64 

93 

25 

46 

71 

38 

13 

57 

d. 72 

26 

47 

94 

55 

32 

17 

63 

86 

e. 31 

15 

67 

83 

22 

45 

96 

54 

95 

f 85 

73 

56 

49 

16 

65 

97 

28 

34 

g. 59 

33 

84 

78 

14 

27 

48 

99 

61 

h. 77 

98 

81 

52 

39 

12 

23 

69 

46 

i. 11 

82 

75 

36 

68 

91 

51 

42 

29 


Read the note at the beginning of Exercise 4. 

1. From each number of row a, subtract: 

a. 3 b. 7 c. 4 d. 8 e. 5 f. 6 g. 9 

2-9. Use rows b, c, d , etc., in like manner. 

10-18. Subtract the numbers given in Example 1 from 
columns a, b, c, etc. 


EXERCISE 13 


Read the note at the beginning of Exercise 4. 


a 

b 

c 

d 

e 

a. 82765 

49038 

76209 

53782 

69458 

b. 75638 

48527 

72589 

51683 

67465 

c. 72947 

43782 

70693 

49785 

63746 

d. 68755 

39476 

69837 

45846 

60983 

e. 65682 

33852 

64967 

41762 

57876 

/. 57765 

28080 

58925 

36547 

52897 

g. 52325 

27863 

45786 

24763 

50079 

h. 49765 

26793 

42687 

21875 

48869 

i. 43689 

19847 

37786 

16954 

38745 

j- 32976 

10756 

29989 

7653 

10594 


INTEGERS 


19 


1. (Parts a, b, c, d, e.) Place a strip of paper below row 

b. Upon it write the results of subtracting the numbers 
in row b from those in row a. 

2. (Parts a, b , c, d, e.) Write similarly the differences 
between the numbers of row c and those of row b. 

3. Subtract the numbers of any row from those in the 
preceding row. 

4. Subtract the numbers of any row from those of 
any preceding row: e. g., those of row e from those of row 
a, etc. 

5. A dentist’s clerk submits to him the following 
record of the accounts of patients whose payments are 
overdue. 


Patient 

Amt. Due on First of Year 

Amt. Paid During Year 

A . . . . 

$ 37.75 

$12.00 

B . . . . 

18.50 

3.75 

C ... . 

132.00 

35.00 

D ... . 

97.25 

23.50 

E ... . 

62.50 

15.00 

F ... . 

85.40 

22.75 

G ... . 

124.60 

33.50 

H . . . . 

106.75 

29.25 


Determine : 

a. the total amount due at the beginning of the year; 

b. the total amount paid during the year; 

c. the amount still due from each patient; 

d. a satisfactory method of checking the accuracy of the 
work. 












20 


NEW HIGH SCHOOL ARITHMETIC 


6 . The total number of bales of cotton raised in the 
Southern States in 1929 was 14,547,791, and in 1930 was 
13,755,518. How many more bales were raised in 1929 than 
in 1930? 

7 . The total amount of money in savings banks in the 
United States in 1930 was $10,357,161,000; and in 1931 
was $ 11,039,910,000. How much was added to the savings 
deposits in the year 1931 ? 

8. The total population of the United States in 1920 
was 105,710,620; in 1930 it was 122,775,046. What was 
the increase in the population during the 10 years from 1920 
to 1930 ? 

9. In 1928 there were 12,703,525 boys and 12,476,171 
girls in.the public schools of the United States. How many 
more boys than girls were there? 

10 . The total area of Texas, the largest state of the 
United States, is 265,896 sq. mi. The area of Germany is 
181,723 sq. mi.; of Austria, 32,369 sq. mi.; of Belgium, 
11,755 sq. mi. How much more than the combined areas 
of Germany, Austria, and Belgium is the area of Texas ? 

11 . Copy and complete the following table which shows 
savings on articles bought at a sale. 


Article 

Former Price 

Sale Price 

Saving 

Tooth paste .... 

$ .50 

$ .29 


Face cream .... 

1.25 

.89 


Face powder 

1.75 

1.23 


Antiseptic .... 

.98 

.69 


Mineral oil ... 

1.00 

.73 


Hand lotion . . . , 

1.00 

.79 


Total . 





Copy and complete each of the tables that appear on 
pages 21 and 22. 















INTEGERS 


21 


12 . A table comparing the attendance in six grades of 
one school during last year and this year. 


Attendance 


UBADE 

Last Year 

This Year 

Increase 

7 

189 

215 


8 

175 

201 


9 

148 

160 


10 

134 

156 


11 

98 

115 


12 

75 

82 


Total 





13. Inventory of a coal dealer for one month. 


Kind 
of Coal 

On 

Bought 

Total 

Sold 

On Hand 

Hand 

IN 

Supply in 

during 

End of 

Mar. 1 

Mar. 

Mar. 

Mar. 

Month 

Egg . . . 

53 

132 


147 


Stove . . . 

112 

245 


310 


Pea . . . 

93 

176 


215 


Lump . . . 

253 

1245 


1162 


Pocahontas . 

173 

648 


579 

• 

Total . 







14. Record of the accounts of five depositors in a bank. 


Depositors 

Total 

Deposits 

Total 

Withdrawals 

Balance 
in Bank 

Mr. Ames . . . 

Mrs. Wilson . . 

Miss Camden . . 

Mr. Bliss . . . 

Mr. Kent . 

$543.75 

168.43 

207.95 

469.32 

307.34 

$126.50 

32.65 

125.15 

278.51 

169.48 


Total .... 











































22 


NEW HIGH SCHOOL ARITHMETIC 


15. A doctor’s record of bills due on the first of a month, 
paid during the month, and due at the end of the month. 


Patient 

Due on 

Jan. 1 

Paid in 
January 

Amount 

Still Due 

Arnold .... 

$ 57.75 

$12.00 


Bennett . . . 

48.50 

13.75 


Casper .... 

102.00 

85.00 


Davis .... 

87.50 

33.50 


Evans .... 

162.50 

45.00 


French .... 

85.40 

32.75 


Gales .... 

124.60 

35.00 


Hancock . . . 

96.75 

40.00 


Total .... 





16. An inventory of several articles in a store. 


Article 

Value op Article 

On Hand 
Mar. 1 

Bought 
in Mar. 

Total 

Sold 
in Mar. 

Remaining 

Fountain Pens . . 

$85.50 

$35.75 


$75.25 


Pencils .... 

25.80 

15.45 


17.55 


Note Books . . . 

63.65 

28.25 


43.35 


Stationery . . . 

54.35 

12.75 


22.65 


Post Cards . . . 

15.28 

6.55 


12.70 


Total . 







8. Counting Out Change is based upon the adding-up 
method of subtraction. 

Example. Eind the change to be given when a $ 5.00 bill 
is offered in payment for purchases amounting to $ 1.63. 

Solution. $1.63 and 2^ is $1.65; and 10 ^ is $1.75; and 25^ 
is $ 2.00; and $ 3.00 is $ 5.00. 

Hence the change consists of two pennies, a dime, a quarter, 
and three $ 1.00 bills; all together, $ 3.37. 































INTEGERS 


23 


EXERCISE 14 

Determine the coins and bills of largest denomination 
with which to “ make change ” when a $ 5.00 bill is offered 
as payment for purchases of: 

1. 33/ 5. 67 / 9. $4.45 13. $1.69 

2. 72/ 6. 23/ 10. $3.87 14. $1.35 

3. 94/ 7. 85/ 11. $2.78 15. $1.94 

4. $1.06 8. $1.15 12. $.22 16. $3.62 

Similarly, when $ 10.00 is offered after the purchase of 
goods that cost: 

17. $2.56 19. $6.89 21. $8.32 23. $2.71 25. $4.58 

18. $4.75 20. $9.27 22. $3.43 24. $5.29 26. $7.86 

Determine the amount and kind of change when the 
buyer offers : 

27. $5.00 after he makes purchases amounting to 18/, 
37/, and $1.00. 

28. $2.00 after he makes purchases amounting to 13/, 
67/, and 70/. 

29. $ 10.00 after he makes purchases amounting to $ 1.25, 
$2.60, $3.15, and 45/. 

30. $ 5.00 after he makes purchases amounting to $ 3.17, 
14/, 29/, and 83/. 

31. $ 20.00 after he makes purchases amounting to $ 9.89, 
$2.57, and $1.98. 

32. $10.00 after purchases amounting to 76/, $2.53, and 
45/. 

33. $5.00 after purchases amounting to 98/, 63/, and 
24/. 

34. $10.00 after purchases amounting to $1.35, $2.65, 
and $5.20. 


24 


NEW HIGH SCHOOL ARITHMETIC 


35. $2.00 after purchases amounting to 28/, 37/, and 
46/. 

36. $ 20.00 after purchases amounting to $ 12.69, $ 3.46, 
82/, and $1.78. 

Multiplication of Integers 

9. In Multiplication, a Multiplier and a Multiplicand are 

given, and a product is to he determined. 

The Multiplicand is the number multiplied ; the Multiplier 
is the number by which the multiplicand is multiplied. 

When the multiplier and multiplicand are integers, the 
product is obtained by using the multiplicand as an addend 
as many times as there are units in the multiplier. 

The symbol for multiplication is x. It will be read 
“ times ” in this book. Hence 3x5 means 5 + 5 + 5 or 15. 

10. If a number of things of one kind be multiplied by 
an integer, the product is things of the same kind. 

Thus, 6 times 7 books is 42 books. 

It is agreed that the multiplier is always a pure or ab¬ 
stract number. Thus 5 cannot be multiplied by 4 ft., nor 
can 6 bu. be multiplied by 7 bu. 

11. Parentheses, ( ), placed about two or more numbers 
which are connected by mathematical symbols, means that 
the result obtained by performing the computation within 
the parentheses is to be treated as a single number in fur¬ 
ther computation. 

Thus, a. 3 + (5 -f 6) means that 5 + 6 or 11 is to be added to 3 

5. 4 x (7 + 8) means that 7 + 8 or 15 is to be multiplied by 4- 

12. Three Fundamental Laws of Multiplication. 

a. Since 3x5 = 5 + 5 + 5 = 15, 

and 5 x 3 = 3 + 3 + 3 + 3 + 3 = 15, 
clearly 3 x 5 = 5 x 3. 


INTEGERS 


25 


This is an illustration of 

The Commutative Law of Multiplication 

The multiplier and multiplicand in a product may be inter¬ 
changed without changing the value of the product. 

b. Since (3 x 2) x 5 = 6 x 5 = 30, 

and 3 X (2 x 5) = 3 x 10 = 30, 
clearly (3 x 2) x 5 = 3 x (2 x 5). 

This is an illustration of 

The Associative Law of Multiplication 

To obtain the product of three integers , the third may be 
multiplied by the product of the first two , or the product of the 
last two may be multiplied by the first. 

c. Since 3 x (4 + 5) = 3 x 9 or 27, 
and (3 x 4) + (3 x 5) = 12 +15 or 27, 

clearly 3 x (4 + 5) = (3 X 4) + (3 X 5). 

This is an illustration of 

The Distributive Law of Multiplication 

To multiply the sum of two (or m,ore) numbers by a number , 
multiply each of them by the multiplier and add the products 
so obtained. 

Example 1. 5 x (2 + 4 + 7) = (5 X 2) + (5 X 4) + (5 x 7) 

= 10 + 20 + 35 
= 65. 

Another form of the distributive law is illustrated in the 
following example. 

Example 2. (3 + 7)x4 = (3x4) + (7x4) 

= 12 + 28 
= 40. 


26 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 15 

1. Prove that 5 x 4 = 4 x 5. 

Of what law is this an example ? 

2. Find in two ways the value of: 

a. 5x(3 + 4) e, 7 X (3 + 2) 

5. 6x (2 + 5) d. 8 x (5 + 4) 

3. Find in two ways the value of : 

a. 5x3x4 c. 3x4x7 

b. 2x4x6 d. 6x2x5 


e. 10 x (6 + 4) 
/. 15 x (8 + 5) 


e. 8x3x4 
/. 9x4x5 


13. Skill in Multiplication depends upon accurate and 
rapid recognition of the products of the forty-five pairs of 
integers on p. 5. Ability to give also the product when 
such integers as 12, 13, 14, and 15 are multiplied by any 
of the integers 1, 2, 3, ••• 9, is well worth securing, and 
ambitious pupils will endeavor to acquire it. 


EXERCISE 16 


Multiplication Drill Table 



a 

b 

c 

d 

e 

/ 

9 

a. 

13 

67 

38 

82 

45 

97 

74 

b. 

26 

83 

65 

92 

66 

29* 

38 

c. 

504 

708 

320 

623 

786 

509 

275 

d. 

3025 

4068 

5409 

6725 

8392 

7653 

8207 

e. 

5746 

7295 

6327 

4008 

8659 

4767 

9724 


Read the note at the beginning of Exercise 4. 

1. Multiply each number of row a by : 

a. 2 b. 3 c. 4 d. 5 e. 6 /. 7 g. 8 ft. 9 

2-5. Use rows b, c, d, and e in the same manner. 

6 - 12 . Multiply all the numbers of the table by 13, 24, 
35, 46, 57, 68, and 98 respectively. 


INTEGERS 


27 


13. Give the result of multiplying each number in row 
a by: 

a. 10 b. 100 c. 1000 

14. Give a rule for multiplying an integer by: 

a . 10 6. 100 c. 1000 

15. Multiply each of the numbers 

3000 240 5300 650 8200 7600 

by: 

a. 2 b. 7 c. 30 d. 50 e. 80 

14. The Multiplication Process is based upon the Distribu¬ 
tive Law of Multiplication. (§ 12.) 

Example 1. Multiply 462 by 7. 

The actual process is : The solution is usually 

7 x 462 = 7 x (400 + 60 + 2) arranged thus: 

= 2800 + 420 + 14 462 

= 3234. _7 

3234 

7x2 units is 14 units, or 1 ten and 4 units; 4 is written in the 
units’ place, and 1 ten is added to the next product. 

7x6 tens is 42 tens. Adding 1 ten gives 43 tens, or 4 hundreds 
and 3 tens. 3 is written in the tens’ place, and 4 hundreds are 
added to the next product. 

7x4 hundreds is 28 hundreds. Adding 4 hundreds gives 32 
hundreds or 3 thousand, 2 hundred. These are written in their 
proper places. 

Example 2. Multiply 65 by 27. 

The actual process is The solution is usually 

27 x 65 = (20 + 7) x 65 arranged thus : 

= (20 x 65) + (7 x 65) 65 

= 1300 + 455 _27 

= 1755. 455 

130 

1755 


28 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 17 


Multiply the following: 


1. 873 by 956. 

2. 2600 by 3950. 

3. 487 by 8009. 

4. 4067 by 997. 

5. 3476 by 625. 

6. 6872 by 599. 

7. 60,507 by 3784. 


8. 15,063 by 9874. 

9. 54,189 by 7998. 

10. 7677 by 4912. 

11. 2946 by 5335. 

12. 82,821 by 7269. 

13. 93,247 by 2461. 

14. 35,895 by 6927. 


15. Eind the value of a hay stack estimated to contain 
35 tons of hay, if hay sells at $ 22 per ton. 

16. A farmer has 145 bushels of potatoes from a 1-acre 
field. What is their value if potatoes are selling at 55^ 
per bushel ? 

17. What is the value of the corn harvested from a 20- 
acre field if the average yield is 32 bushels per acre and the 
crop sells for 37 ^ per bushel ? 

18. What is the value of the wheat harvested from a 
20-acre field if the average yield is 19 bushels per acre and 
the crop sells for 65 $ per bushel? 

19. What is the value of a crop of oats harvested from a 
20-acre field if the average yield is 45 bushels per acre and 
the crop sells for 29 $ per bushel ? 

20. What is the cost of 675 school desks if each desk 
costs $4.85? 

21. What is the cost of furnishing schooling to 2385 
pupils at $ 95.45 each ? 

22. A certain electric light company makes a charge of 
26^ per foot for extensions of its line. 

a. What will be the cost for running a line 692 feet ? 

b. What will be the cost for running a line four miles ? 

c. What will be the cost for running a line 175 rods ? 


INTEGERS 


29 


23. A farmer has fifteen, cows, each of which yields, on 
the average, 6200 pounds of milk per year. What is his 
income from the sale of milk if he gets $ 1.23 per hundred 
pounds ? 

24. A florist has the following plants on hand to he sold 
at the indicated average prices during the Christmas holi¬ 
days. 

500 plants to sell at $ 2.50 800 plants to sell at $ 1.25 

500 plants to sell at $ 1.75 1250 plants to sell at $ .65 

What will he receive if he sells all of them ? 

25. In a certain city, 77 pensioners received payments 
from the firemen’s pension fund, as follows: 

29 widows who received $ 20 a month each; 

28 minors who received $ 6 a month each; 

13 disabled firemen who received $ 50 a month each; 

5 retired firemen who received $ 40 a month each; 

2 dismissed firemen who received $ 30 a month each. 

What was the total payment per month from the fund ? 

Short Methods of Multiplication 

15. There are several Short Methods of Multiplication that 
can be used so often as to justify an effort to learn them. 

a. Multiplying by 9 is the same as multiplying by 
( 10 - 1 ). 

Thus: 9 x 56 = (10-1) x 56 

= (10 x 56) - (1 x 56) 

= 560 - 56 
= 504. 

The solution may be arranged thus: 

56 _ 560 

_9_ ” 56 

504 

Rule. — To multiply any number by nine, annex a zero to it 
and from the result subtract the given number. 


30 


NEW HIGH SCHOOL ARITHMETIC 


Example. Multiply 463 by 9. 

Solution. 463 _ 4630 

9 “ 463 

4167 

b. Multiplying by 99, 999, etc., may be performed in sim¬ 
ilar manner j e.g. 

Rule. — To multiply any number by 99, annex two zeros to 
it (i. e. multiply by 100) and then subtract the given number. 

Example. Multiply 234 by 99. 

Sdutim. 9 9 x 234 = 2 ^ g ° 4 0 

23166 

c. Multiplying by 50 is the same as multiplying by \ of 
100. Hence, 

Rule. — To multiply any number by 50 annex two zeros to 
it (i. e. multiply by 100) and divide the result by 2. 

Example. Multiply 436 by 50. 

Solution. 436 x 50 = -43600 - 21800. 

d. Multiplying by 25 is the same as multiplying by ^ of 
100. Hence, 

Rule. — To multiply any number by 25, annex two zeros 
to it and then divide by 4. 

Example. Multiply 789 by 25. 

Solution. 789 x 25 = JZJ ? o o « 19725. 

e. Multiplying by 15 is the same as multiplying By 
(10 + 5). But the product of any number by 5 is one half 
its product by 10. Hence, 

Rule. — To multiply any number by 15, first annex a zero 
to the number and then to the result add one half itself. 






INTEGERS 


31 


Example. Multiply 234 x 15. 

Solution. 234 x 15 = 2340 = 10 x 234 
1170 = i of 2340 
3510 

/. Multiplying by 75 is the same as multiplying by 
(50 + 25). But the product of any number by 25 is one 
half its product by 50. Hence, 

Rule. — To multiply any number by 75, annex to it two 
zeros and divide by 2. (See Rule (c)), and then add to this 
result one half itself. 

Example. Multiply 378 by 75. 

Solution. 378 x 75 = 2137800 

18900 = 50 x 378 
9450 = \ of 18900 
28350 

g. Rules (a) and ( b ) are only special cases of the device 
illustrated in the following examples: 

Example 1. Multiply 76 by 103. 

Solution. Since 103 = 100 + 3, 

103 x 76 = (100 + 3) x 76 

=(100 x 76)+(3 x 76) 

= 7600 + 228 
= 7828 

The solution should be arranged thus: 

103 x 76 = 7600 = 100 x 76 
228 = 3 x 76 
7828, adding 

Example 2. Multiply 237 x 98. 

Solution. 98 = 100 — 2. 

.-. 98 x 237 = (100 x 237) - (2 x 237). 

The solution should be arranged thus : 

98 x 237 - 23700 = 100 x 237 
474 = 2 x 237 

23226, subtracting 





32 


NEW HIGH SCHOOL ARITHMETIC 


This form of multiplication can be employed to advantage 
whenever the multiplier is a number a little more or a little 
less than 10, 100, 1000, etc. 

h. Familiarity with these rules and careful scrutiny 
of the multipliers will suggest other short methods of 
multiplication. 


EXERCISE 18 

1. Multiply each of the following numbers by 9 : 

a. 27 c. 78 e. 263 g. 496 i. 624 k. 463 

b. 46 d. 94 /. 375 h. 547 j. 938 l. 758 

2. Multiply each of the numbers given in Example 1 
by 99. 

3. Multiply each of the numbers given in Example 1 
by 999. 

4 . Multiply each of the numbers given in Example 1 
by 50. 

5. Multiply each of the numbers given in Example 1 
by 25. 

6. Multiply each of the numbers given in Example 1 
by 15. 

7. Multiply each of the numbers given in Example 1 
by 75. 

8. Multiply the first six numbers in Example 1 by 97. 

9. Multiply the first six numbers in Example 1 by 102. 

10. Multiply the first six numbers of Example 1 by 98. 

11. Multiply the first six numbers of Example 1 by 105. 

12. Find the cost of 75 caps at 99^ each. 

13. Find the cost of 325 bushels of corn at $1.04 per 
bushel. 

14. Find the cost of 137 bushels of potatoes at 93 1 per 
bushel. 


INTEGERS 


33 


15. Eind the total length of 76 hoards, each 25 inches 
long. 

16. Eind the total weight of 130 bags of sugar, each 
weighing 50 pounds. 

17. Eind the cost of 56 window shades at 75^ each. 

18. Eind the cost of 23 pounds of beef at 25 i per pound, 

19. Eind the cost of 35 heads of lettuce at 15 $ per head. 

20. Eind the cost of three dozen shovels at $ 1.50 each. 

21. Eind the cost of 54 bags of cement at 50 ^ per bag. 

22. If concrete curbing and gutter on a street cost 65 ^ 
per running foot, what is the cost for the curbing and 
gutter on both sides of a city block 625 feet long? 

23. Eind the cost of 745 bushels of corn at 50 $ per 
bushel. 

24. Eind the cost of excavating 1345 cubic yards of dirt 
at 25 $ per yard, 

25. How much should a farmer receive for 78 pounds of 
spring chicken at 25 $ per pound ? 

26. How much pay is due an employee who has worked 
54 hours at the rate of 50 $ per hour ? At the rate of 25 ^ 
per hour? At the rate of 75^ per hour? At the rate of 
15 $ per hour ? 

27. The cashier in a lunch room takes in three 15^ 
checks, seventy-four 25^ checks, and thirty-three 50^ 
checks. How much money was taken in ? 

28. The ticket seller at a game sold 428 tickets at $1.25; 
642 at $ 1.50 ; and 233 at $ 1.75. How much cash should 
he have ? 

29. What is the cost of 13 yards of lace at 94^ per yard ? 

30. How much should a fruit grower receive for 37 cases 
of strawberries at $ 2.10 per case ? 

31. In making a 16-ft. concrete highway, 100 bags of 
cement were used for each 50 feet. What is the total num¬ 
ber of bags of cement required for 1 mile of highway ? 


34 


NEW HIGH SCHOOL ARITHMETIC 


Invoices 

16. An Invoice is an itemized statement of articles or 
goods bought and their cost. (Most people not engaged in 
business call an invoice a bill.) 

A Statement is an itemized report from one person or 
firm to another for work done, and the charge made for it. 


Northtown, Illinois. 
Feb. 18, 1933. 

Johnson Coal Company, 

Two Lakes, Wisconsin. 


Bought of The Northern Illinois Mining Company. 
Terms. Cash in 30 days. 


2/15/33 

37 T 

#2 Lump Coal 

. $2.35 

$ 86 

95 



2/17/33 

65 T 

Mine Run . , 

. $1.89 

122 

85 

$209 

80 


$ 86.95 is found by multiplying $ 2.35 by 37. 

Filling in the amounts $ 86.95, $ 122.85, and the total 
$ 209.80 is called Making the Extensions. This is probably 
a new phrase to you. Fix it in mind now. 

Sometimes the following form of invoice is used. 

AGRANO NURSERY CO. 

South Winthrop, Illinois 


Sold to Bender Floral Co. 

Washington, Indiana 

Terms. 60 days. Date 3/23/33 


450 

Persian Lilac . . . 

. . 7.5* 

$33 

75 



350 

Hungarian Lilac . . 

. . 12.5* 

43 

75 

$77 

50 























INTEGERS 


35 


EXERCISE 19 

Prepare the following bills. Use last year in dates. 

1. On Sept. 23, the Davis Fnr Co. sold to A. M. Johnson 
Co., Owasee, Ill., the following items: 

3 pair gloves #4345 @ $5.25; 1 pair gloves #4475 
@ $6.75; 12 pair gloves #4375 @ $2.60; 8 pair boys’ 
mittens #3450 @ $ .79; and 9 pair gloves #9354 @ $4.20. 

2. On March 15, the Jenkins Saw Co. of Indianapolis, 
Indiana, sold to the Marshville Hardware Co., Marsh- 
ville, Indiana, the following goods : 

i doz. 26" ripsaws #8 m 320 @ $1.65; i doz. 28" rip¬ 
saws #8 m 320 @ $1.75; doz. 26" handsaws #8 m 315 
@ $ 1.58; i doz. 24" handsaws # 8 m 330 @ $ 1.95; 1 doz. 
22" panel saws #8 m 332 @ $1.83; 1 doz. 26" handsaws 
#8 m 340 @ $2.15; -J doz. 28" ripsaws #8 m 342 @ $2.45; 
and i doz. hacksaw outfits #8 m 350 @ $ 1.78. 

3. The Central Book Co. of Detroit, Mich., on Aug. 24, 
sold to J. C. Carpenter, Stationer and Bookseller, Detroit, 
Mich., the following books : 

24 copies Corn Cookery by Cade @ 60 /; 8 copies Mod¬ 
ern Cook Book by Jewett @ $ 1.35; 15 copies Simple 
Cookery by Tyson @ $ 1.00; 25 copies Meals Without 
Meat by James @ $ 1.80; 36 copies Wheatless Meals by 
Morgan @ 63 /. 

4. The States Auto Supply Co. of Chicago, Ill., sold on 
April 17 the following goods to the Northern Indiana 
Garage, Crown Point, Indiana: 

1 doz. wheel pullers @ 22 / each; 2 doz. rear tire holders 
@ $1.75 each; 3 doz. sets cork gaskets @50/ per set; 
i doz. steel autolocks @ $1.70 each; 11 doz. pump con¬ 
nections @ 25 / each; 4 doz. boxes repair patches @ 
$8.40 per dozen; 8 Merkel vulcanizers @75/ each; 15 
Ford timers @ 85 / each; 8 doz. Chic spark plugs @ 35 / 
each; 18 two-cylinder pumps @ $ 1.35 each. 


36 


NEW HIGH SCHOOL ARITHMETIC 


5 . John Dahl, Carpenter, Maumee, Ill., on May 17, sub¬ 
mitted the following bill for labor and materials furnished 
for a repair job for Andrew Cameron: 

400 ft. oak flooring @ $ 85 per 1000 ft.; 400 ft. hemlock 
flooring @ $ 40 per 1000 ft.; 2 doors @ $ 5.35; 100 ft. 
finishing oak @ $ 120 per 1000 ft.; 5 (8 hr.) days master 
carpenter @ 60 ft per hr.; 5 days journeyman carpenter 
@ 45 $ per hr.; 5 days apprentice @ 30 ^ per hr. 

6. The Browning Shoe Co., East St. Louis, Ill., sold to 
A. M. Jeffrey, Pardee, Ill., the following goods, on May 15 : 

10 pr. shoes #1507 @ $3.73; 8 pr. shoes #1208 @ 
$3.56; 15 pr. shoes #1397 @ $4.00; 5 pr. shoes #1505 
@ $6.70; 20 pr. shoes #1061 @ $3.20; 25 pr. oxfords 
#1303 @ $2.45; 15 pr. oxfords #6241 @ $2.90; 24 pr. 
slippers #1393 @ $1.35; 18 pr. brown shoes #1340 @ 
$ 4.45 ; and 6 pr. brown shoes # 1807 @ $ 4.25. 

7 . On Feb. 25, the Pay son Brush Co. of Cleveland, Ohio, 
sold to Morgan Bros., Interior Decorators, the following 
goods: 

i doz. 8" whitewash brushes #131 @ $2.50; i doz. 
8i" whitewash brushes #131 @ $2.95; f doz. 7" kal- 
somine brushes #310 @ $1.25; § doz. 8" kalsomine 
brushes #310 @ $1.43; i doz. 7" kalsomine brushes 
#311 @ $1.72; J doz. 8" kalsomine brushes #311 @ 
$2.12; 1| doz. 8" paste brushes #331 @ $2.32; f doz. 
10" smoothing brushes #335 @ $1.97; 1| doz. 12" 
smoothing brushes #335 @ $2.43; 1\ doz. dusting 

brushes #321 @ $.63. 

8. The Western Table Co. of Grand Rapids, Mich., 
sold on March 10, to Kellog and Burgess, Furniture 
Dealers of Kalamazoo, Mich., the following goods : 

25 kitchen tables #L240 @ $2.15; 6 porcelain-top 
tables #L243 @ $6.78; 18 drop-leaf tables #L241 @ 


INTEGERS 


37 


$ 7.42 ; 5 extension tables ft L 250 @ $ 11.78 ; 12 extension 
tables ft L 253 @ $ 14.55 5 3 dining tables ft L 248 @ 
$ 17.67 ; 4 dining-room tables #L 257 @ $ 23.87. 

9. Martin and Lewis, Wholesale Clothiers, New York 
City, sold on Jan. 30, to the Hub Clothing Store of Jersey 
City, New Jersey, the following goods : 

15 boys’ suits ft 3700 @ $10.70; 18 boys’ suits #1300 
@ $10.85; 24 boys’ suits #1031 @ $9.75; 25 boys’ suits 
ft 9300 @ $ 9.85 ; 12 boys’ suits ft 6300 @ $ 12.35. 

10 . Mr. Otto Niemann built for Mr. William Harvey on 
his estate a drive, and on Sept. 17 submitted a bill for the 
following items: 

Grading — 3 men, 2 days each @ $2.75 per day; 1 man, 
2 days @ $4.00 per day; 2 teams, 2 days @ $5.50 per 
day. Stone, and spreading stone — 96 cu. yd. crushed 
stone @85^ per yd.; 42 cu. yd. screenings @ 75 ft per yd. 
Labor — 2 men 3 days each @ $2.75 per day; 1 man 3 
days @ $4.00 per day. Steam roller — 1 day @ $10.00 
per day. 

Division of Integers 

17. In Division two numbers are given, the Dividend and 
the Divisor. Division is the process of determining the 
number, called the Quotient, by which to multiply the di¬ 
visor so as to give the dividend as product. The symbol 
for division is -s-. 

Thus, 15 3 is 5, for 5x3= 15. 

The dividend is the number divided; the divisor is the 
number by which the dividend is divided. 

If the dividend and divisor are integers and the quotient 
is also an integer, the division is exact; then, 

dividend = quotient X divisor. 

When the dividend and divisor are integers, and no 
integer can be found which will serve as quotient, then the 


38 


NEW HIGH SCHOOL ARITHMETIC 


division is not exact, and there is a Remainder. This re¬ 
mainder is the difference between the dividend and nearest 
smaller integer which does exactly contain the divisor. 

Thus, 17 does not exactly contain 3. 15 is the next smaller 

integer which does exactly contain 3. 17 — 15 or 2 is the re¬ 

mainder when 17 is divided by 3. Notice that 

17 = 5 x 3 + 2. 

In general, dividend = quotient x divisor -f- remainder. 

18 . The following forms for division are suggested : 

Example 1. Divide 756 by 3. 

252 

Explain this solution. 

Example 2. Divide 22236 by 68. 

68 is contained in 222 three times with a re¬ 
mainder. Place 3 in the quotient above the 2 in 
the hundreds’ place. 3 x 68 = 204. Subtracting 
204 from 222 gives 18. Annex 3 to 18, forming 183, 
183 contains 68 twice with a remainder. Place 
2 in the quotient in the tens’ place. 2 x 68 = 136. 
Subtracting 136 from 183 gives 47. Annex 6 to 47, 
forming 476. 476 contains 68 seven times. Place 
7 in the quotient in the units’ place. 7 x 68 = 476. Subtract¬ 
ing 476 from 476, the result is zero. Hence the quotient is 327, 
and the remainder is zero. 


327 

68122236 

204 

183 

136 

476 

476 


3J756 


Example 3. 

209 
386[80791 
772 
3591 
3474 


Divide 80791 by 386. 


In this solution, after subtracting 
3474, there is the remainder 117. 


117 = Remainder 








INTEGERS 39 


EXERCISE 20 
Division Drill 


a 

b 

c 

d 

e 

/ 

9 

a. 56 

78 

104 

93 

187 

216 

1302 

b. 48 

97 

233 

75 

128 

463 

3522 

c. 39 

86 

326 

85 

209 

527 

2674 

d. 66 

45 

448 

67 

382 

307 

4849 

e. 96 

59 

267 

44 

175 

438 

5367 


Read the note at the beginning of Exercise 4, p. 5. 

1 . Give the quotient and the remainder when each num¬ 
ber of row a of the foregoing table is divided by: 

a. 2 b. 3 c. 4 d. 5 e. 6 /. 7 g. 8 h. 9 

2-5. Use rows b, c, d, and e in the same manner. 

Note. — Other drill examples can be based upon this table. 
For example, divide the numbers of the table by 11, 12, 13, etc. 

EXERCISE 21 


Division Drill Table 


a 

b 

c 

d 

e 

a. 8557 

11266 

12212 

14749 

21414 

b. 10260 

16059 

22032 

31698 

38826 

c. 10792 

18164 

27968 

37012 

56088 

d. 24795 

39498 

49068 

60204 

74472 

e. 29406 

47236 

61247 

73138 

85940 


1 . Divide the numbers of row a by 43. 

2 . Divide the numbers of row b by 54. 

3. Divide the numbers of row c by 76. 

4. Divide the numbers of row d by 87. 

5 . Divide the numbers of row e by 98. 

Note. — Other drill examples can be made by changing the 
divisors. For example, divide every number in the given table by 
215. 


40 NEW HIGH SCHOOL ARITHMETIC 

19 . In applications of division, two types of problems 
occur. 

If any number of things of one hind be divided by another 
number of things of the same hind , the quotient is an abstract 
number. 


Example 1. A train travels 35 miles an hour. In bow 
many bonrs will it travel 105 miles if it travels always at 
tbe same rate ? 

Solution. 1. 105 miles 35 miles = 3. 

2. Hence it will take 3 hours. 

If any number of things of one hind be divided by an 
abstract number , the quotient will be a number of things of the 
same hind as the dividend. 

Example 2. A train travels 120 miles in three bonrs. 
How far does it travel in one hour if it travels always 
at the same rate ? 


Since it travels 120 miles in 3 bonrs, it will travel in one hour 
as many miles as 3 is contained times in 120, or 40 miles. 

120 h- 3 = 40 miles. 

In applications of division, tbe remainder frequently baa 
no meaning and must be ignored. 


Example 3. An antoist stopping to buy gasoline finds 
that be has just $ 1.67. If gasoline is selling for 22 / per 
gallon, bow many gallons may be ask for ? 


Solution. $ 1.67 -r- $ .22 = 7+. 

That is, be has enough money for 7 gallons, but 
not enough for 8 gallons. He buys 7 gallons and 
has 13 f left over. 


7 

22J167 

154 

13 


This example illustrates one of tbe most common forms 
in which division is used. Tbe division is performed with 
tbe expectation that there will be a remainder, tbe desire 
being to find tbe largest possible integral quotient. 


INTEGERS 


41 


EXERCISE 22 

1. How many windows can be glazed with 144 panes of 
glass if four panes are required for each window ? 

2. How many carloads of 40 tons each will be required 
to deliver 33,600 tons of coal ? 

3. Electric light companies place poles between 100 ft. 
and 120 ft. apart. Assuming that the poles are placed 
110 ft. apart, how many poles are required for one mile of 
electric light line? If the poles cost $ 22.50 each, what is 
the total cost for poles ? 

4. How many bed sheets 99 inches long can be cut from 
a bolt of muslin containing 50 yards, if 4 inches must be 
allowed on each sheet for hems ? 

5. In 1930, about 14,243,000 bales of cotton were raised 
in the United States, and sold for about $ 674,044,000. 
What was the average price per bale ? 

6. A long ton of American coal occupies about 43 cubic 
feet. How many tons in round numbers can be carried in 
a vessel having a storage capacity of 180,000 cubic feet ? 

7. How many kitchen towels about 30 in. long can be 
cut from a 10-yard bolt of toweling, no allowance being 
made for hemming ? 

8. How many window curtains each 68 in. long can be 
cut from 36 yards of curtain material, if 4 in. are allowed 
on each curtain for the hem at bottom and top ? 

9. How many pillow slips 30 in. long can be cut from a 
bolt of pillow-slip tubing containing 10 yards, if 4 in. are 
allowed for hems on each slip ? 

10. A man decides to spend about $ 200 for coal. If 
coal costs $ 11.50 per ton, how many full tons of coal may 
he order ? 


42 


NEW HIGH SCHOOL ARITHMETIC 


11 . If oats are selling for $ .33 per bushel, how many 
bushels may a man order who cannot spend more than 
$ 10.00 ? How much change will he receive from his 
$10.00 bill? 

12. How many shrubs at $ .35 each may a man order 
who can afford to spend at most $ 15.00 ? 

13. If a maid can save $ 3.75 a week, how many weeks 
will it take her to save enough to buy a suit costing 
$22.50? 

14. A worker folding surgical dressings folded 75 in 
one hour. What was the average time for folding each 
dressing ? 

15. If an automobile travels one mile in 45 seconds, how 
many miles will it travel in an hour at that rate ? 

16. The following ration, approximately, is used by a 
certain agricultural college for feeding its dairy cows : 


Silage.3 lb. per day per 100 lb. live weight 

Hay.1 lb. per day per 100 lb. live weight 

Beets.1 lb. per day per 100 lb. live weight 


Concentrated feed . . . 1 lb. per day per 3| lb. of milk pro¬ 

duced per day 

How many tons, approximately, of silage, hay, beets, 
and concentrated feed are required for a herd of 10 cows 
that have an average weight of 1050 pounds and produce 
on the average 7000 pounds of milk per year per cow? 

17. The deposits of eight depositors of a bank on a cer¬ 
tain day were: $ 428.25, $ 68.34, $ 125.30, $ 552.75, $ 79.84, 
$277.56, $274.50, and $161.38. What was the average 
of their deposits ? 

18. The sales of five departments in a store one month 

were: $9560.25, $7820.15, $7253.35, $3745.30, and 

$ 9823.60. What was the average of their sales ? 





INTEGERS 


43 


19. In 1929 there were 14,919,000 bales of cotton raised 
in the United States. They were sold for approximately 
$ 1,225,032,000. What was the average sale price per bale ? 

Note. When dividing, you can drop the three zeros from the 
divisor and the dividend. 


20. Complete the following table, which gives some 
statistics about the production of tobacco in the United 
States in 1930. 


State 

Acres 

Pounds Raised 

Average Yield 
per Acre 

Massachusetts 

8,100 

11,397,000 


Wisconsin . . . 

43,000 

52,900,000 


North Carolina . 

779,000 

535,195,000 


Kentucky . . . 

507,500 

331,700,000 



21. On the 62 square miles of the District of Columbia, 
there were living in 1930 486,869 persons. How many 
persons was that per square mile ? 


Factoring Integers 

20. A Prime Integer is one that can be divided exactly 
only by itself and unity; as, 1, 2, 3, 5, 7, 11, etc. 

A Composite Integer is one that can be divided exactly by 
other integers as well as itself and unity; as, 4, 6, 8, 9, 
10, etc. 

21. Every integer by which a number can be divided 
exactly is a Factor of that number. 

Thus, 3, 6, 8, 12, etc., are factors of 24. 

22. An Even Number is one that can be divided by 2. 

23. The Prime Factors of an integer are the prime integers 
whose product is the given integer. 

Thus, the prime factors of 24 are 2, 2, 2, and 3, for 
2x2x2x3 = 24. 










44 


NEW HIGH SCHOOL ARITHMETIC 


24. To Factor an integer means to find its prime factors. 

Rule. — To factor an integer : 

1. Divide the integer by one of its prime factors. 

2 . Divide the quotient by one of its prime factors. 

3 . Continue in this way until a prime quotient is obtained. 

4 . The complete set of factors consists of the various divisors 
and the final quotient. 

Example. Factor 1092. 

Solution. 1. 1092 = 2 x 546 

2. = 2 x 2 x 273# (for 546 = 2 x 273) 

3. =2x2x3x91 (for 273 = 3 x 91) 

4. =2x2x3x7x13 (for 91 = 7 x 13) 

5. the factors of 1092 are 2, 2, 3, 7, and 13. 

In applying this rule, always start with the smallest 
prime integer that appears to be a factor. Also, use the 
form of solution employed in the illustrative example. 

25. The following Rules of Divisibility aid in selecting 
the factors of a number. 

I. A number is divisible by 10 if it ends in zero. 

II. A number is divisible by 5 if it ends in zero or five. 

III. A number is divisible by 2 if it ends in zero or an even 
number. 

IV. A number is divisible by 3 if the sum of its digits is 
divisible by three. 

Thus, 384 is divisible by 3 since 3+8 + 4 or 15 is divisible by 3. 

The proofs of these rules appear in § 30, p. 46. 


INTEGERS 


45 


EXERCISE 23 

1. What are the prime numbers between 1 and 100 ? 
Find the prime factors of the following numbers ; 


2. 

26 

8. 

75 

14. 

212 

20. 

310 

26. 

738 

3. 

38 

9. 

84 

15. 

237 

21. 

345 

27. 

828 

4. 

42 

10. 

96 

16. 

256 

22. 

382 

28. 

885 

5. 

54 

11. 

110 

17. 

264 

23. 

444 

29. 

962 

6. 

63 

12. 

128 

18. 

275 

24. 

526 

30. 

1000 

7. 

68 

13. 

145 

19. 

294 

25. 

612 

31. 

1256 


26. A Common Divisor or Common Factor of two or more 
integers is a factor of each of them. 

Thus, 2 is a common divisor of 18, 24, and 38. 

27. Two numbers that do not have a common divisor ex¬ 
cept 1 are prime to each other. 

Thus, 8 and 11 are prime to each other. 

28. The Greatest Common Divisor (G. C. D.) of two or more 
integers is the greatest integer that is a divisor of each of 
them. 

Thus, 6 is the G. C. D. of 18, 24, and 42. 

29. Finding the G. C. D. of Given Integers. 

Example. Find the G. C. D. of 144, 264, and 540. 
Solution. 1. Factor each of the integers. (§ 24) 

144 = 2x2x2x2x3x3 
264 = 2 x 2 x 2 x 3 x 11 
540 = 2x2x3x3x3x5 

2. 2 and 3 are the only common factors. 

3. 2 appears only twice as a factor in 540. 

3 appears only once as a factor in 264. 

The G. C. D. = 2 x 2 x 3 = 12. 


4. 


46 


NEW HIGH SCHOOL ARITHMETIC 


Rule. — To find the G. C. D. of two or more integers : 

1. Find the prime factors of each of them. 

2 . Form the product of all the different prime factors, using 
each of them the least number of times that it occurs as a 
factor in any of the given numbers. 


EXERCISE 24 


Find the G.C.D. of: 



1 . 

36 and 144. 

8. 

72, 60, and 84. 

2. 

108 and 132. 

9. 

60, 105, and 150. 

3. 

81 and 126. 

10. 

36, 90, and 126. 

4. 

165 and 210. 

11. 

117, 81, and 126. 

5. 

288 and 648. 

12. 

104, 182, and 351. 

6. 

128 and 192. 

13. 

525, 3375, and 7425. 

7. 

675 and 1125. 

14. 

160, 224, and 260. 

15. A farmer has three pieces of timber whose lengths 
are 63, 84, and 105 feet, respectively. What is the length 
of the longest logs all of the same length that can be cut 


from them? 


16. Two schools, containing 480 and 672 pupils, respec¬ 
tively, are divided into classes containing the same number 
of pupils. What is the greatest number of pupils that can 
be taken for a class ? 

SUPPLEMENTARY TOPICS 

30. A. Proofs of the Rules of Divisibility. (Page 44) 

Proof of I. — A number whose last digit is 0 consists of a 
number of tens. Hence any number ending in 0 can be di¬ 
vided by 10. 

Thus, 560 consists of 56 tens, for 500 = 50 tens and 60 = 6 tens. 


INTEGERS 


47 


Proof of II. —If a number ends in 0, it consists of a num¬ 
ber of tens. Each ten can be divided by 5. Hence the 
whole number can be divided by 5. 

Thus, 360 = 36 tens, and therefore 72 fives. 

If a number ends in 5, it consists of a number of tens and 
five units. 

Thus, 365 consists of 36 tens and 5 units. The tens are divis¬ 
ible by 5 and also the 5 units are divisible by 5. Hence the whole 
number is divisible by 5. 

Proof of III. — If a number ends in 0, it consists of a num¬ 
ber of tens. Each ten is divisible by 2. Hence the number 
is divisible by 2. 

If a number ends in 2, 4, 6, or 8, it consists of a number 
of tens plus 2, 4, 6, or 8 units. 

The tens are divisible by 2, and also the units are divis¬ 
ible by 2. Hence the whole number is divisible by 2. 

Proof of IV. — Consider the number 684. 

1. 684 = 6 hundreds + 8 tens + 4 units 

= 6(99 + 1) + 8(9 + 1) + 4 
= 6x99 + 6 + 8x9 + 8 + 4 
= 6x 99 + 8x9 + (6 + § + 4) 

2. But 6 x 99 and 8x9 are each divisible by 3, since 99 and 9 
are divisible by 3. 

3. Therefore 684 is divisible by 3 if (6 + 8 + 4) is divisible by 3. 

4. 6 + 8+ 4 = 18 = 6x3. 

5. Therefore 684 is divisible by 3. 

B. Greatest Common Divisor by a Division Process. 

31. Finding the G. C. D. of Two or More Numbers that Can¬ 
not be Factored Easily. 

Example 1. Eind the G. C. D. 221 1493 |2 
of 221 and 493. 442 

Solution. 1. Dividing 493 by 221, the 51 

quotient is 2 and remainder is 51. 

2. Any factor of 221 is also a factor of 442, since 442 = 2 x 221. 




48 


NEW HIGH SCHOOL ARITHMETIC 


3. Hence any factor common to 493 and 221 mnst also be com¬ 
mon to 493 and 442, and therefore to 493 and 51. 

4. Also, any factor common to 221 and 51 must occur in the 
result when 51 is added to 2 x 221; i.e., it mnst occur in 493. 

5. Hence 221, 493, and 51 have the same common factors. In 
general, — the G. C. D. of any two integers is the same as the G. C. D. 
of the less integer and the remainder obtained when the greater is di¬ 
vided by the less. 

6. Dividing 221 by 51, we have 

7. ,\ the G. C. D. of 51 and 221 is the same as 
the G. C. D. of 17 and 51. 

8. Dividing 51 by 17, the quotient is 3. 

Hence the G. C. D. of 51 and 17 is 17. 

9. Therefore 17 is the G. C. D. of 51 and 221. * (Step 7.) 

Therefore 17 is the G. C. D. of 221 and 493. (Step 5.) 


51[221] 
204 
17 


This solution illustrates the 

Rule. — To find the G. C. D. of two integers : 

1. Divide the greater by the less. 

2 . If there is a remainder, divide the previous divisor by it. 

3 . Continue in this manner, making each remainder a divisor 
of the preceding divisor until there is no longer a remainder. 

4 . The last divisor is the G. C. D. of the given integers. 


Example 2. Find the G. C. D. of 377 and 667. 

Solution. 1. 377 [667] 1 

377 

290 [377] 1 
290 

87 [290] 3 
261 

29[87]3 

87 

2. Therefore the G. C. D. of 377 and 667 is 29. 

Note. — If the G. C. D is 1, the integers are prime to each 
other. (§ 27) 



INTEGERS 


49 


EXERCISE 25 

Find the G. C. D. of the following sets of integers: 


1. 559 and 817. 

2. 391 and 598. 

3. 589 and 899. 

4. 703 and 893. 

5. 533 and 1271. 

6. 731 and 1247. 

7. 3658 and 4602, 


8. 3703 and 6923. 

9. 1591 and 2183. 

10. 5605 and 6785. 

11. 6059 and 7446. 

12. 5312 and 10043. 

13. 2291 and 3713. 

14. 10057 and 11659. 


Least Common Multiple of Integers 

32. A Multiple of an integer is any integer which will 
contain the given integer as an exact divisor. 

Thus, 12, 20, 32, etc., are all multiples of 4. 

33. A Common Multiple of two or more integers is a mul¬ 
tiple of each of them. It will contain each of them as an 
exact divisor. 

Thus, 36 is a common multiple of 2, 3, and 9 ; also 18 and 54 
are common multiples of 2, 3, and 9. 

34. The Least Common Multiple of two or more integers 
(L. C. M.) is the smallest integer that will contain each of 
them exactly as a divisor. 

Thus, the L. C. M. of 2, 3, and 9 is 18. 

35. Finding the L. C. M. of Given Integers. 

Example 1. Find the L. C. M. of 40, 84, and 144. 

Solution. 1. Factor each of the given integers. (§ 24) 

40 = 2x2x2x5 
84 = 2x2x3x7 
144 = 2x2x2 x 2x3x3 

2. The L. C. M. must contain 144 and hence it must contain 
the factors 2, 2, 2, 2, 3, and 3. 

3. In order to contain 84, the L. C. M. must contain the factors 
of 84; that is, the factors 2, 2, 3, and 7. It must therefore con¬ 
tain the factor 7 besides those named in Step 2. 


50 


NEW HIGH SCHOOL ARITHMETIC 


4. In order to contain 40, the L. C. M. must contain the factor 
5, as well as the factors 2, 2, 2. 

5. Therefore the L. C. M. must have the factors 2, 2, 2, 2, 3, 3, 
5,7. 

That is, the L. C. M. = 2x2x2x2x3x3x5x7 = 5040. 

Rule. — To find the L. C. M. of two or more integers : 

1. Find the prime factors of each of the integers. 

2 . Form the product of all the different prime factors that 
appear in Step 1, using each of them the greatest number of 
times that it occurs in any one of the given integers. 

Example 2. Find the L. C. M. of 24, 60, and 105. 

Solution . 1. 24 = 2x2x2x3 


60 = 2x2x3x5 
105 = 3 x 5 x 7 


2. 2 appears as a factor three times in 24, and less frequently 
in the other integers. 

3 appears as a factor only once in any of the integers. 

5 appears as a factor once in 60. 

7 appears as a factor once in 105. 

3. .-. the L. C. M. = 2 x 2 x 2 x 3 x 5 x 7 = 840. 

Note. — If the L. C. M. can be seen by inspection, then cer¬ 
tainly get it that way, — do not take this longer way. Thus, the 
L. C. M. of 2, 3, and 4 is evidently 12. 


EXERCISE 26 


Eind the L. C. M of: 


1. 4, 6, and 9. 

2. 3, 5, and 10. 

3. 6, 7, and 8. 

4. 8, 10, and 15. 

5. 4, 14, and 21. 

6. 4, 8, and 20. 

7. 3, 6, and 15. 

8 . 8, 12, and 18. 

9. 6, 9, and 12. 
10. 18, 38, and 54 


11. 36, 104, and 24. 

12. 33, 44, and 55. 

13. 108 and 144.- 


14. 20, 75, and 180. 

15. 32, 88, and 121. 

16. 28, 49, and 147. 

17. 15, 35, and 77. 

18 . 52, 81, and 117. 

19. 98, 126, and 140. 

20. 115, 138, and 230. 


II. FRACTIONS 


36. A Common Fraction consists of two integers, a Numera¬ 
tor and a Denominator. It is usually expressed by writing 
the numerator over the denominator, separated by a hori¬ 
zontal line. The numerator and the denominator together 
are called the terms of the fraction. 

In elementary arithmetic, the denominator is regarded as 
indicating a number of parts into which a whole is divided 
and the numerator as indicating the number of the parts 
being considered. 

Thus, f is interpreted as three “ one fourths ” of a unit. 

Mathematicians regard a fraction as indicating the di¬ 
vision of the numerator by the denominator. 

Thus, -§• is interpreted by them as 3 divided by 4, or one fourth of 
three equal units. 

Both meanings are useful in applications of fractions. 

The two meanings are equivalent, in the case of integers. 

In the following figure, rectangle AB represents a unit 
and AC represents 3 units. Study of the figure discloses 
that AD is three “ one fourths ” of the unit, AB, and is also 
one fourth of the three units, AC. 


i of AC 



c 


l of AB 


37. A Proper Fraction is a common fraction whose numera¬ 
tor is less than its denominator ; as -§. 

An Improper Fraction is a common fraction whose numera¬ 
tor is greater than its denominator; as J. 

A Mixed Number is the sum of an integer and a fraction; 
as 5 -f- f, or, as it is usually written, 5f. 


51 




52 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 27 


1. Illustrate geometrically the fraction -J. 


What kind of 

fraction is this ? 

2. Illustrate geometrically the fraction f. What kind of 
fraction is this ? 

3. Illustrate geometrically the number 2\. What kind of 
number is this ? 

4. Illustrate geometrically the fact that | of 2 is the 
same as f of 1. 

38. Two Fundamental Laws of Fractions Verified. 

In the figure below, let rectangle AG represent one unit. 
Let it be divided into six equal parts AB, BC, CD, DE, 
EF, and FG. Then AC, CE, and EG represent three equal 
parts of AG. 

A B c D E F G 

1 


^or| 1 

Hence AE represents f of AG and also of AG. 

Hence f- = f. 

Notice that f can be obtained from f by multiplying 
both numerator and denominator by 2. 

Rule I. — The numerator and denominator of a fraction may 
each be multiplied by the same number without changing the 
value of the fraction. 

Referring again to the fact that -§ = £, or that | = |, 
notice that J can be obtained from £ by dividing both 
numerator and denominator by 2. 


Rule II. — The numerator and denominator of a fraction 
may each be divided by the same number without changing the 
value of the fraction. 





FRACTIONS 


53 


EXERCISE 28 

1. a. Give a geometrical illustration of the fact that 
J = \. b. By what are each of the terms of i multiplied to 
give | ? 

2. By what are the terms of f multiplied to give f ? 

3. Multiply both numerator and denominator of f by : 

a. 2 b. 3 c. 5 d. 10 e. 15 

How do all the resulting fractions compare ? 

4. Divide both numerator and denominator of -J$ by : 

a. 10 b. 5 c. 2 

How do all the resulting fractions compare? 

5. Write five fractions each equal to the fraction f, and 
tell how each is obtained from f. 

6. Write five fractions each equal to the fraction ■£, and 
tell how each is obtained from 

7. Write three fractions each equal to the fraction 
and having smaller terms than it. How is each obtained ? 

8. Write two fractions each equal to the fraction -ffc, 

9. Change each of the following fractions to an equal 
fraction having a smaller denominator and tell how you 
do it: 

a. |f b. T \ c. if d. U e. ff 

/• tVV g- « <• « J- t 8 A 

10. Change each of the following fractions to an equal 
fraction having a larger denominator and tell how you do it: 
a. f b. | c. f d. i e. § f £ 

9- | h. % i. j. ^ f 4 

39. A fraction is in its Lowest Terms when its numerator 
and denominator do not have any common factor. 


54 


NEW HIGH SCHOOL ARITHMETIC 


40. Reducing a Fraction to its Lowest Terms. 

Example 1. Reduce ff to its lowest terms. 

Solution. Dividing both numerator and denominator by 7, 

35 _ 5 . 

6^—9 

Example 2. Reduce to its lowest terms. 

Solution. 1. Dividing both numerator and denominator by 25, 

112 5 — .45 
2 7 0 0 — 1 08 

2. Dividing both terms by 9, - 5 ^. = 


. 1125 _ 5 

**2700 12 


3. 


The solution should be arranged thus : 



Rule. — To reduce a fraction to its lowest terms : 

1 . Divide both terms by any common factor of them. 

2 . Divide both terms of the resulting fraction by any of their 
common factors. 

3 . Continue in this manner until there results a fraction 
whose terms do not have a common factor. 

Note 1. — This process is equivalent to dividing both terms by 
their G. C. D., so a fraction can be reduced to lowest terms at once 
by dividing both terms by their G. C. D. 

Note 2. — Special care should be taken when the last remain 
ing factor in the numerator is 1 , as in the following example. 

Example 3. Reduce Y 3 T °o' t° its lowest terms. 


1 


Solution. 


3 

30 _ W _ 1 


210 ?X0 7 * 


n 

7 


Example 4. Reduce to its lowest terms 


21 x 12 x 15 
20x14 x 36* 






FRACTIONS 


55 


Solution. 


1 

3 13 

?lxZ2xl3 _ 3xlxl _3 
w x n X 33 4x2x1 8 

4 2 3 

1 


In this solution, 21 and 14 are each divided by 7, and the 
quotient 3 is written above 21 and 2 below 14. 12 and 36 are 

each divided by 12, and the quotient 1 is written above 12 and 3 
below 36. 15 and 20 are each divided by 5, and the quotient 3 is 

written above 15 and 4 below 20. The 3 above 15 and the 3 below 
36 are each divided by 3, and the quotients 1 are written down. 

The final result is the product of the remaining factors in the 
numerator over the product of the remaining factors in the 
denominator. 

Note. — The process of dividing out common factors of a 
numerator and a denominator is commonly called cancellation. It 
is preferable to think of it as a process of division. 


EXERCISE 29 


Express in their lowest terms the fractions : 



21 . ~ . 
55 x 112 


49 x 88 


26. 


8 x 12 x 14 X 15 
10 x 16 x 18 x 21* 


22 

' 7 x23 x 30* 
00 20 x 108 


34 x 38 x 39 
* 57 x 85 x91* 


23 s ' w 

18x28 x 36* 


16 x 95 x 96 
114x128 
21 x 39 x 55 


29. —. 

39 X 56 x 117 

54 x 84 x 270 
* 50x162x 196* 


27 x 77 x 105 
135x165 
21 X 26 x 52 


25 ^ ^ 

' 20x26x33 














56 


NEW HIGH SCHOOL ARITHMETIC 


31. A set of gimlets consists of one of each of the follow- 

luo* Q17PQ *2 3 4 5 6 7 8 9 1 0 1 1 1 2 
m 5 bizeb . 3^? T2> 3~2> 3~2> TO TO TO TO TO TO TO 

Express each of these sizes in its lowest terms. 

32. A set of auger bits consists of one of each of the fol¬ 
low! no- • 4 5 7 8 9 10111213141516 

lowing SlZCb . y-g, yy, j g’j yg-, yg, yg-, yg-, -yg, T g, yy, -yg, Jq. 

Express each of these sizes in its lowest terms. 


41. Changing an Improper Fraction to an Integer or a Mixed 
Number. 


Example 1. Express as an integer. 

Since ^ = 54 4 - 9, = 6. 


Example 2. Express ^ as a mixed number, 


^ = 45 + 3 = 5+ 3= 53 



Rule. — To express an improper fraction as an integer or a 
mixed number: 

1 . Divide the numerator by the denominator. 

2 . The fraction equals the quotient, or, if there is a remain¬ 
der, the quotient plus the fraction formed by writing the re¬ 
mainder over the divisor. Reduce the fractional part of the 
result to lowest terms if necessary. 


Example 3. How many bushels of oats are there in 
1578 pounds of oats if one bushel of oats weighs 32 pounds ? 


Solution. The number of bushels = 18 

= 4914 


49 


32|1578 
128 




FRACTIONS 


57 


EXERCISE 30 

Express each of the following fractions as an integer or 
a mixed number: 


1 . 

f ^5. 


5. gal. 

9. 

- 1 3 ^°- gross. 

2. 

-¥ yd- 


6 * fff 

10. 

bu. corn. 

3. 

f|in. 


7 £4 in 

/. 16 1U. 

11. 

-2^°- bn. wheat. 

4. 

¥ P k - 


Q 6400 T 

2 0 00 

12. 

- 2 j^ oz. 


13. 

38 

V 

19. 

25. 

if- 

31. 

Hi 1 -- 

14. 

¥• 

20. 

-W- 26. 

2 7 9 

~s r- 

32. 

H¥-- 

15. 

45 

TT* 

21. 

W* 27. 

59 7 

tv- 

33. 


16. 

5 6 
~9~* 

22. 

W- 28. 

1194 

~22~- 

34. 

**F* 

17. 

¥• 

23. 

29. 

2167 

78~- 

35. 

2 3 6 9 

18. 

9 2 

12* 

24. 

-196 30 

16 * 

2 7 8 6 * 

36. 

W* 


42. Expressing a Fraction as an Equal Fraction Having a 
Larger Denominator. 


Example 1. Express j- as an equal fraction with de¬ 
nominator 28. 


Solution. 

2 . 


1. 28 -s- 4 = 7. 

Multiply both terms of the fraction by 7. 

3 _ 7 x 3 _ 21 

4 7 x 4 28* 


Then 


Rule. — To express a fraction as an equal fraction having 
a larger denominator: 

1. Divide the new denominator by the given denominator. 

2 . Multiply both terms of the fraction by the quotient. 

Example 2. Express 22 as a fraction with denominator 

12 . 

Solution. 1. 22 = \ 2 -. 

2 . 12 - 1 = 12 . 

22 _ 12 x 22 _ 264 
“1 12 x 1 12 * 





58 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 31 

1. Express § as an equal fraction having the denominator: 

a. 12 b. 9 c. 15 d. 24 e. 36 /. 60 

2. Express f as an equal fraction having the de¬ 
nominator : 

a. 12 b. 20 c. 36 d. 48 e. 60 /. 56 

3. Express | as an equal fraction having the de¬ 
nominator : 

a. 16 b. 24 c. 40 d. 56 e. 72 /. 96 

4. Express f as an equal fraction having the de- 
nominator: 

a. 15 b . 25 c. 35 J d. 45 e. 100 /. 120 

5. Express £ as an equal fraction having the de¬ 
nominator : 


a. 

12 b. 24 


c. 

36 

d. 

48 

e. 

96 

/. 72 

6. 

Express 6: 

a. 

as 

8ths 

b. 

as 

5ths 

c. 

as lOths 

7. 

Express 8 : 

a. 

as 

9ths 

b. 

as 

4ths 

c. 

as 7ths 

8. 

Express 9: 

a. 

as 

5ths 

b. 

as 6ths 

c. 

as 8ths 

9. 

Express 11: 

a. 

as 3ds 

b. 

as 

5ths 

c. 

as lOths 

10. 

Express 13: 

a. 

as 

4ths 

b. 

as 

6ths 

c. 

as llths 


43. Changing Two or More Fractions to Equal Fractions 
having their Lowest Common Denominator. 

To add two fractions like i and i, it is necessary to 
change i to | and i to f. Then 

i+i=f+i=f 

| and f have a common denominator. Until fractions do 
have a common denominator, they cannot be added or 
subtracted. 

The lowest common multiple of the denominators of the 
fractions is used as the lowest common denominator of the 
fractions. 


FRACTIONS 


59 


Example 1. Express and as equal fractions 

having their lowest common denominator. 


Solution. 

1. The L. C 

. M. of 6, 

10, and 15 is 

; 30. 

2. For f: 

i 

o 

CO 

6 = 5. 

5 _ 

5 x 5_ 

.25 




6 

5x6 

'30* 

For : 

+ 

o 

CO 

10 = 3. 

7 _ 

3x7 

_ 21 



10 

3 x 10 

30* 

For t 4 3 : 

30 -T- 

15 = 2. 

4 _ 

2x4 

_ 8 




15 

2 x 15 

30* 


Rule. — To express fractions as equal fractions having their 
lowest common denominator: 

1. Find the L. C. M. of their given denominators. (§ 35) 

2 . For each fraction, divide the L. C. D. by the given denomi¬ 
nator, and then multiply the numerator and the denominator of 
the fraction by the quotient. 


EXERCISE 32 

Change the following fractions to equal fractions having 
their lowest common denominator. 


1. b 

l 

2' 


5. 

5 

6’ 

b 

q 3 
y * 8> 

2. 

5* 


13- 

A 

2 * 

3 

T* 


6. 

3 

4* 

10. f, 

1 

2* 


14- I, 

3 

¥• 

3. 1, 

I- 


7. 

3 

1 

ii- i, 

2 

¥* 


15- |, 

4 

¥* 

4. b 

2 

¥* 


8. 

b 

1 

3* 

J.-S. y, 

5 

■6* 


16- 

7 

¥* 

17. 

b 

b and f 



24. 


H 

, and yj. 


18. 

b 

b and |. 



25. 

3 

1 6> 

A 

, and f. 


19. 

5 

6> 

T 7 g-, and | 



26. 

T2> 


and -J. 


20. 

31 

24 

2 1 

’ 32> 

and 

25 

48‘ 


27. 

5 

3 

8? 

and T 7 g-. 


21. 

3 13 

1 0> TT> 

and 

22 

¥¥* 


28. 


5 

and -fe. 


to 

to 

T2 

9 

j 2in 

and 

H- 


29. 

A? 


and 


CO 

ea 

11 15 

l¥> T8> 

and 

& 


30. 



and i|. 





60 


NEW HIGH SCHOOL ARITHMETIC 


44. Adding Fractions. 


Example 1. 
Solution. 1. 
denominator 12. 

2 . 


Find the sum of -f, f, and f. 

Express each fraction as an equal fraction with 



+ t 


T2 


= n= i=2|. 


(§ 35 ) 


Rule. — To add two or more fractions : 

1. Express them, if necessary, as equal fractions having their 
lowest common denominator. (§ 43) 

2 . Add the numerators of the resulting fractions for the 
numerator of the sum, and write the result over their L. C. D. 

3 . Reduce the resulting fraction to its lowest terms. 

Mixed numbers are added by adding separately the in¬ 
tegral parts and the fractional parts. 


Example 2. 

Find the sum of 3 J, 

4|, and 5|. 





Solution. 1. 3f + 4| + 5-| = 

3a + 

ns + 





2. 



= 

12H- 






3. But 


f I — 

or 2j. 





4. 



12H = 

12 + 2i 

, or 14b 








EXERCISE 33 






Find the sums 

: 








1. 

I + i 

2^4* 

6. 

3. 4_ 1 

8^4* 

11. 

3 _i_ 5 

4 + T2‘ 

16. 

3 

¥ 

+ 

5 

6* 

2. 

1 _L 1 
¥ + 3* 

7. 

2. 1 5. 

3^6* 

12. 

1- -1- - 9 - 
8 ' l¥* 

17. 


+ 

b 

3. 

1 i 1 

4 + 3- 

8. 

3. I 7 
¥ ' ¥' 

13. 

4 i 3 

3+1- 

18. 

I 

+ 

b 

4. 

* + f 

9. 

1-1-3 

3 ' 8’ 

14. 

5 i 3 

2 "T f 

19. 

5 

8 

+ 

5 

6- 

5. 

1 + f- 

10. 

3. -L ! 

2 i 6* 

15. 

f+l- 

20. 

1 

8 

+ 

f 

21. 

i + A + A- 


25. 

1 + A+! 

3 

T* 




22. 

A + A + 

2 

5* 


26. 

A + 3¥ + 

1 4 
5T* 




23. 

7 ,i_ 11 _L_ 1 

¥ • T3 ' 3T* 


27. 

A + M + 

2.3 




24. 

A + 2 T + 

I* 


28. 

f+ 4 r + x 7 s 

f* 





FRACTIONS 


61 


29 - A + A + A- 

30 - A + A + if 

31. lf + lf + lf. 

32. li+2f + 3 T V 

33. 7f + .31 + If. 

34. 3| + 5| + *. 

Note. — If other examples a 
in the examples of Exercise 32, 


35. 6i + 5| + 4|. 

36. 3i + 9i + 7f. 

37. 4 + 3f + 2 r V 

38. 6f + 2+8£. 

39. -f + 2i 

40. 7i + 2f+3|. 

re wanted, add the sets of fractions 
p. 59. 


41. The manager of a restaurant ordered three whole 
hams. When delivered, they weighed 10J, 11 §, and 12^ 
pounds respectively. What was the total weight ? 

42. Five workmen, A, B, C, D, and E, had at the end of 
a week the following record of hours worked each day of 
the week preceding. 


Mon. 

Tues. 

Wed. 

Thttrs. 

Fbi. 

Sat. 

A. 8 




8f 

4 

B. 8 

8 

H 

n 

8 

H 

C. 8i 

8 

8 


9 

Bi 

D. 8f 

8 

8f 


H 

Bi 

E. 8 

ihN 

00 



8 

4f 


How many hours did each man work during the week ? 

43. The figure at the right is a cross section 
of an iron gutter to be made by a tinsmith. It 
is made by bending a piece of material of the 
proper width. How wide must the material be ? 

44. A small drawing board is built up of three layers of 
wood. Two layers are in thickness and one is 
What is the total thickness of the board ? 



45. A veneered oak door has its stiles and rails built up 
as follows : there is a piece of quartered oak veneer T V f in 
thickness on both sides of a piece of plain oak l-fa" in 
thickness. What is the total thickness of the rails and 
stiles ? 


62 


NEW HIGH SCHOOL ARITHMETIC 


46. What will be the total height of a set of sectional 
bookcases consisting of a base that is 10" high, two sec¬ 
tions 12^" high, one section 10|-” high, and a top 3" high ? 


I 


Plaster 


Studding 




i Lath. 
Studding 
=? Lath 


47. The adjoining figure repre¬ 
sents a section through a partition 
wall in a dwelling. 

If the studding is 3|" wide, the plaBter 

lath f" thick, and the plaster thick, how thick is the 
partition ? 


48. The adjoining figure represents a section through 
an outside wall of a 
stuccoed house. 

If the studding is 
3J" wide, the lath f" 
thick, the inside plaster 

thick, the sheathing f " thick, the furring strips f " thick, 
and the stucco f" thick, what is the total thickness of 
the wall ? 



Lath 


45. Expressing a Mixed Number as an Improper Fraction. 

Example 1. Express 2-J as an improper fraction. 

Solution. 1. 2f means 2 + f. 

2. But 2 = -f, and therefore 2£ = £ + £ = J£L. 

Rule. — To express a mixed number as an improper frac¬ 
tion : 

1. Multiply the integral part by the denominator of the 
fractional part, and to the product add the numerator of the 
fractional part. 

2 . Write the result of Step 1 over the denominator of the 
fractional part. 

Example 2. Express 9| as an improper fraction,, 

Solution. 1. 8 x 9 = 72; 72 + 7 = 79. 




















FRACTIONS 


63 


EXERCISE 34 

Express each of the following as an improper fraction. 


1 . 

31- 

11. 

00 

£ 

21. 

4*- 

31. 

9f. 

2. 

SI- 

12. 

10f 

22. 


32. 

12f. 

3. 

4f- 

13. 

6f. 

23. 

^1T* 

33. 

17f. 

4. 

6f- 

14. 

5f 

24. 

3fV- 

34. 


5. 

4f 

15. 

3f 

25. 


35. 

40 J - 
-±^1 6 . 

6. 

5f. 

16. 

12f. 

26. 

14f. 

36. 

Sfi- 

7. 

7f- 

17. 

18f. 

27. 

9|- 

37. 

27f 

8. 

9f. 

18. 

15f 

28. 

8*. 

38. 

66f. 

9. 

10J. 

19. 

21J. 

29. 

10|- 

39. 

71f 

10. 

12f. 

20. 

4f- 

30. 

13tV 

40. 

96f 


46. Subtracting Fractions. 

Example 1. Subtract A from -f. 

Solution. | — tt — if — = uV 

Rule. — To subtract one fraction from another : 

1. Reduce the fractions if necessary to their L. C. D. 

2 . Subtract the numerator of the subtrahend fraction from 
that of the minuend, and write the result over the L. C. D. 

3 . Reduce the final result to lowest terms. 

Example 2. Subtract 2-J from 5f. 

Solution. 5| — 2| = 5f^ — = 3^. 

Rule. — To subtract mixed numbers: 

1 . Express the fractional parts of the mixed numbers in 
terms of their L. C. D. 

2 . If the fractional part of the subtrahend is less than that of 
the minuend, subtract the integral parts and the fractional 
parts separately. (See Example 2.) 


64 


NEW HIGH SCHOOL ARITHMETIC 


3 . If the fractional part of the subtrahend is greater than 
that of the minuend, add 1 from the integral part of the minu¬ 
end to the fractional part of it, and then subtract as in Step 2 
of this rulec (See Example 3.) 

Example 3. Subtract from 5J. 

Solution. 1. — 3| = 5-j^g- — 

2. = m - Hi (5* = 4ft.) 

3. = 1H 


EXERCISE 35 

Eind the value of: 


1. 

1 _ 

2 

1 

S' 

11. 

*- 

“ i- 

to 

H 

5- 

3 

~S' 

31. 

2®A - 

-5J. 

2. 

2 _ 
■3 

1 

1’ 

12. 

A “ 

_ 3 

S' 

22. 

7 — 

4 

T* 

32. 

33f- 

12|. 

3. 

i ~ 

1 

3‘ 

13. 

tf “ 

_ 3 

4' 

23. 

10- 

_ 5 

S'- 

33. 

99f — 

2f. 

4. 

3 _ 
5 - 

1 

¥* 

14. 

TO " 

_ 7 

8* 

24. 

6- 

2 

¥* 

34. 

36 y — 

7f- 

5. 

i- 

2 

3’ 

15. 

A ~ 

_ 2 

~5' 

25. 

8 - 

5 

9* 

35. 

18|- 

6yV 

6. 

7 _ 

8 

f 

16. 

tf- 

_ 8 
■8* 

26. 

12 - 

-f 

36. 

22} — 


7. 

5 _ 

6 

t* 

17. 

tt- 

_ 5 

T6- 

27. 

5 i- 

“ 

37. 

131|- 

-128|. 

8. 

7 _ 

8 

. 5 

IT* 

18. 

T2 - 

_ 3 
~S‘ 

28. 

6f- 

_ 2 

3* 

38. 

107}- 

- 105f 

9. 

A " 

_ 3 

4* 

19. 

12 ~ 

- A- 

29. 

*A 

-3i- 

39. 

99| — 

98|. 

10. 

i- 

2 

■3* 

20. 

2 5 
3T 

_ 3 
¥* 

30. 

15 A ~ 21 . 

40. 

114} - 

-109}. 


41. From a barrel containing 55 gal. of gasoline, a farmer 
took 51 gal. for his car. How many gallons of gasoline had 
he left in the barrel ? 

42. From a bolt of cloth containing 151 yd., a clerk sold 
a piece 4| yd. long. How many yards were left in the 
bolt? 


FRACTIONS 


65 


43. From a leg of beef weighing 35f lb., a butcher sold a 
piece weighing 4-J lb. How much did the remainder of the 
piece weigh? 

44. From a board which was 7-J in. wide, a carpenter 
cut a strip l T 9 g in. wide. How wide was the piece which 
remained ? 

45. I have a strip of wood which is 3|- in. wide. How 
wide a strip must I nail on the edge to make the total 
width 4f in. ? 

46. In one end of a piece of „_ 

wood which is f" thick, a slot is ^ 11 

to be cut so that of wood shall remain on each side of 
the slot. How wide will the slot be? 

47. The adjoining figure represents a panel 
door. The outside pieces are each 2f" wide. 

Assume that the panel fits into a groove f" > 

deep as shown by the dotted lines. How 

large must the panel be made ? [ 

Multiplication of Fractions 

47. Multiplying a Fraction by an Integer. 

Example 1. 2xf = f + | = | = f* 

Notice that the final result can be obtained at once by writing 
the numerator over the quotient of 8 divided by 2. 

Similarly, 3 x -g = f, and 5 X ^ = |- 

Example 2. 3xf = f + l + l = f- 

Notice that the final result can be obtained at once by multi¬ 
plying the numerator 2 by 3, and writing the result over the 
denominator 5. 

Similarly, 2 X f = f, and 5 x | = 4r- 

R U l e . — To multiply a fraction by an integer: 

i. When the denominator can be divided by the integer, 












66 


NEW HIGH SCHOOL ARITHMETIC 


write the given numerator over the quotient of the denominator 
divided by the given integer. 

2 . Otherwise, multiply the numerator of the given fraction 
by the integer, and write the result over the given denominator. 
Reduce the result to lowest terms. 


Examples of the second kind are solved rapidly as follows : 

7 3 7 

Example 3. 24 x - = X - = 3 x 7 = 21. 

8 8 


48. Multiplying a Mixed Number by an Integer. 

Example 1. Multiply 3f by 5. 

Solution. 5 x 3f = 15 + ^ = 15 + 3£ = 18£. 

Rule. — To multiply a mixed number by an integer: 

Multiply the integral part and the fractional part separately 
and add the results. 


EXERCISE 36 

1. Multiply each of the following fractions : 



2 3. 5 3 4 
IP 45 65 85 IT5 

TIP ¥5 

b 

11 
12 * 




a. 

by 2 b. by 3 c. by 4 

d. by 5 

e. 

by 8 

f 

by 20 

2 . 

Multiply by: a. 2 

b. 5 

c. 

7 

d. 

10 

e. 13 

3. 

Multiply fZg- by: a. 4 

b. 6 

c. 

11 

d. 

13 

e. 12 

4. 

Multiply 2 ^- by: a. 7 

b. 5 

c. 

14 

d. 

11 

e. 28 


Find each of the following products 


±z 


32* 


6. 45 x f. 

7. 62 x*. 

8. 95 x f. 

9. 60 x 
72 x*. 


10 . 


11. 256 xf. 

12 . 


1432 x if. 


13. -J of 90. 

14. -I of 234. 

15. of 84. 

16. of 642. 


17. f of 965. 

18. f of 1578. 

19. i of 365. 

20. A of 273. 

21. 9x3f. 

22 . 10x5f. 


FRACTIONS 


67 


23. 

12 

X 

6 f- 

29. 

45 x 16|. 

35. 

98 x 66f. 

24. 

14 

X 

H- 

30. 

136 x 81 

36. 

17 x 55J-. 

25. 

21 

X 

3f 

31. 

224 x 142. 

37. 

212 x 4|. 

26. 

65 

X 

si- 

32. 

638 x 11|. 

38. 

64 x 3 t V 

27. 

92 

X 

si- 

33. 

83 x 211. 

39. 

48 x Ilf. 

28. 

24 

X 

12f 

34. 

75 x 331. 

40. 

96 X 2y * 1 2 3 * * * * 2-. 


49. Multiplying a Fraction by a Fraction. 

In the following figure is a geometrical illustration of 

I off 

' ____ F _ B 

A 

Rectangle AF is clearly of rectangle AB. The shaded 
part of AF is -J of AF. 

Therefore this shaded part of AF is -J of 1 of AB. 

It is clear from the figure that this shaded part of AF is 
also yh of AB. 

Therefore f of l of AB is T 8 y 0 f AB. 

Notice that T 8 y can he obtained by writing 2x4 over 

3x5. 

Rule. — To multiply one fraction by another : 

1. Multiply the numerators of the fractions for the numerator 
of the product, the denominators for the denominator of the 
product. 

2 . Reduce the resulting fraction to its lowest terms. 

Thus, 1 X f = and f X f = |f. 

Since a fraction is reduced to lowest terms by dividing 

both numerator and denominator by any common factors of 

them, it is preferable to remove such common factors, as in 

the following: 




68 


NEW HIGH SCHOOL ARITHMETIC 


Example 1. 
Solution. 


Find the value of -f- x -J X 
2 1 

0 S 11 = 22 
7 ? 1^ 63' 

3 3 


In this solution, 6 and 9 are each divided by 3, and 5 and 15 
are each divided by 5, the quotients being written either above or 
below the number divided. The result is obtained by multiplying 
the resulting factors of the numerator for the numerator and 
those of the denominator for the denominator of the product. 

In multiplication examples involving mixed numbers, it is 
usually preferable to change the mixed numbers to improper 
fractions. 


Example 2. Find the value of x 2 T j X 

1 


Solution. 1A- x 2 A x |=^x^x- = —• 
20 9 XV 9 25 


EXERCISE 37 

Find the following products : 


1 . 

f x \K 

8 . 

18. 

2l> 

V 20 

A TT 

X 45 

A 74* 

15. 

3|x£ 

if 

2 . 

txf 

9. 

1 0 
2T 

v 1 2 
X 2"5 

xf. 

16. 

16| 

X 

3?V 

3. 

9 v 11 

16 A 36* 

10 . 

7. y _6 V 1 0 

8 A 49 A 6 3* 

17. 

37! 

X 

4 A- 

4. 

Ax£. 

11 . 

21 

X 55 
A 9T 

X 35 

A ”6 * 

18. 

33! 

X 

xf O'* 

5. 

«xff 

12 . 

1 3 
T2 

V 20 
A 37 

A 47* 

19. 

87! 

X 

2f 

6 . 

18 y 24 

4 2 A ITT* 

13. 

2 2 
T" 

X - 1 # 

X 15. 

20 . 


X 

2 A x i-A 

7. 

«x«. 

14. 

1 4 

27 

x A 

X 5 4 

A T3* 

21 . 


X 

2!fx* 

Find the cost of 









22 . 

12 lb. of sugar at 4| 

t pei 

lb. 





23. 

14 T. of coal 

L at i 

11| per 

T. 






24. 2| lb. round steak at 23 t per pound. 

25. 4$ lb. chicken at 24 / per pound. 


FRACTIONS 


69 


26. 37^ lb. chicken at 18^ per pound. 

27. 13 gal. gasoline at 23 T 2 ¥ $ per gallon. 

28. 85 bu. oats at 29f ^ per bushel. 

29. 275 bu. corn at 36f ^ per bushel. 

30. 1450 bu. # 3 white oats at 22| ^ per bushel. 

31. 750 bu. flaxseed at $ 1.71J per bushel 

32. 2500 bu. wheat at 45f ^ per bushel. 

33. 12,500 lb. peanuts at 3f ^ per pound. 

34. 3250 lb. prunes at 1\$ per pound. 

35. 875 lb. lard at per pound. 

36. 9876 qt. milk at 4 1 ^^ per quart. 

37. 10,965 lb. sugar at 4 T y 7 ^ per pound. 

38. 1375 yd. calico at 4-f P per yard. 

39. 1985 yd. gingham at 5^/ per yard. 

40. 1575 yd. overcoating at $ 1.911 per yard. 

41. 1250 yd. overcoating at $ 1.41f per yard. 

42. 2750 yd. sheeting at 5j^ per yard. 

43. 965 lb. oxide of zinc at 5-| $ per pound. 

44. A meat market owner purchased a whole beef weigh¬ 
ing 448 lb. at 141^ per pound. Also 77 lb. of heavy pork 
loins at 9-J-^ per pound, 68 lb. of medium loins at 9^ per 
pound, and 82 lb. of light loins at lOJ-^ per pound. What 
was the total cost? 

45. Find the cost of 12 T. of bran at $ 17-J per ton. 

46. A recipe for golden corn cake calls for f cup (c.) of 
corn meal, 1^ c. of sugar, 4 teaspoonfuls (t.) of baking- 
powder, i t. of salt, 1 c. of milk, 1 egg, and 1 tablespoon¬ 
ful (T.) of melted butter. 

a. What quantities should be used for one and one-half 
portions of this recipe? 

b. What quantities should be used for two and one-fourth 
portions of this recipe? 



70 


NEW HIGH SCHOOL ARITHMETIC 


47. The following recipe is given for coffee sauce to be 
served with ice cream: 

li c. of milk, i c. of ground coffee, J c. of sugar, f T. of 
arrowroot, and salt. 

How much of each should be used to make one half of 
the recipe? 

48. A recipe for orange snow pudding is : 

1 T. of gelatine, \ c. of cold water, f c. of boiling water, 
•J c. of sug^r, j c. of orange juice^2 T. of lemon juice, and 
the white of 1 egg. 

Determine the quantities to use for one third portion of 
this pudding; also for one and one half portions. 

49. In feeding dairy cows, it is suggested that \ pound 
of concentrated feed (bran, middlings, etc.) be fed for 
every pound of milk produced daily. 

How many pounds of feed should each of the following 
cows receive ? 

Cow # ^ produces 28 lb. per day cow # 2, 30 lb. per day; 
cow jf 3, 25 lb. per day ; cow jf 4, 33 lb. per day. 

50. A recipe for plum preserves calls for three fourths 
as much sugar as fruit, and one cup of water to each pound 
of sugar. How much water and sugar are required for 
14 lb. of fruit ? for 9 lb. ? for 17 lb. ? 

51. In a certain institution, members of the faculty re¬ 
ceive as salary for service during the summer session 0 f 
their annual salary. Determine the amounts to be paid to 
persons whose salaries are: 

a. $1200 b. $1750 c. $2250 

52. What is the cost of 4f T. of coal at $11.70 per ton? 

53. What is the cost of yd. of cloth at 16|^ per 
yard ? 

54. What is the cost of 276 lb. of sugar at 5J^ per 
pound ? 


FRACTIONS 


71 


55. What is the cost of 17-| A. of land at $ 235 per acre ? 

56. Find the cost of 18f ft. of steel rod at $ .35 per foot, 
and 241 ft. of rod at $ .46 per foot. 

57. A machinist is working at the rate of 65^ per hour, 
with “ time and one half ” for overtime. (The full day is 
8 hours.) What should be the amount of his wages at the 
end of a week in which he worked 8 hr. on Monday, 9 on 
Tuesday, 10 on Wednesday, 9^- on Thursday, 8f on Friday, 
and 8 on Saturday ? 

58. A member of a club received the following statement 
of his indebtedness to the club for use of the billiard room: 

May 8, If hr.; May 12, If hr.; May 18, 21 hr.; May 24, 
11 hr.; May 30, If hr. What was the amount of his bill 
at 30 $ per hour ? 

59. Find the cost of 237 bales of silk averaging 125f yd. 
at $ 2.43 per yard. 


Division of Fractions 

50. Dividing a Fraction by an Integer. — To divide a frac¬ 
tion by an integer like 2 means to find the fraction which 
must be multiplied by 2 to produce the given fraction as 
product. 

Thus, f 2 = f, for 2 x f = f. 

Similarly, If -r- 3 = T 4 7 , for 3 X T 4 7 = if. 

Also, | -h 2 = |, for 2 x f = |. 

Similarly, f 3 = r 2 5 , for 3 x T 2 5 = f. . 

Rule. — To divide a fraction by an integer : 

1. If possible, divide the numerator by the integer and write 
the result over the denominator. 

2 . Otherwise, multiply the denominator by the integer and 
write it below the given numerator. 

Thus, - 4 / 5 = f, and | 5 = T V 


72 


NEW HIGH SCHOOL ARITHMETIC 


51. Dividing an Integer or a Fraction by a Fraction. 

Example 1. Divide 3 by §. 

Solution. 1. 3 = A 5 -* 

2. ... 3 -H- | = 15 fifths - 2 fifths = 15 -s- 2 = - 1 /-. 

Notice now that = 3 x so that 3-f | = 3 x | - f is 
said to be the fraction § inverted. 


Example 2. f by •§-. 

Solution. 1. ? ~ and 

O .3.2 _ 9.8 _ Q . Q _ 9 

<£. • • V — 3 — T2 ~ T2 — y “ 0 — S '• 

Notice again that f = | x so that |• -f- f = f x f. The quo¬ 
tient is found by multiplying | by the fraction f inverted. 

Rule. —To divide a number by a fraction : 

1 . Multiply the dividend by the inverted divisor. 

2 . Reduce the result of Step 1 to lowest terms. 


Example 3. Divide -f-f by 


Solution. 


5 3 

12 18 12 35 n 35 14 ' 

2 7 


If mixed numbers appear, change them to improper 
fractions. 


Example 4. 
Solution. 


Divide 13J by 2\. 
131 2* = -V + f. 


27 

2 


x 


3 2 

9 % 9 

1 1 


= 6. 


EXERCISE 38 

Find the value of: 


1. 

T* 

4. 

i 

9 

re- 

7. 

2 5 

3*12* 

2. 

3 _t_ _2_ 

I 1 10* 

5. 

3 

¥ * 

9 

r- 

8. 

1 3 7 

T6 * ¥* 

3. 

5. _s_ 2. 

6 * 3* 

6. 

3 _s_ 

■8 * 

6 

9. 

15. _i_ A 

16 * 





FRACTIONS 



73 

.10. 

a -■ 

■_ 7 

• s- 

22. 

A 

34. 

66f- 

-25. 

11 . 

*- 

A* 

23. 

1-3. 

35. 


-4-12. 

12. 

16- 


24. 

i + 4. 

36. 

ni¬ 

90 

. T . 

13. 

22- 


25. 


37. 

si- 

■m- 

14. 

35- 


26. 

•H-4. 

38. 

11 JL 

9~ • 

■2f 

15. 

24 - 

-t- 

27. 

6|-5. 

39. 

63 • 
52 “ 

9 

T3- 

16. 

32- 

-!■ 

28. 

2* + 3. 

40. 

28- 

42 

TT‘ 

17. 

12- 

L_ 5 
* Q' 

29. 

CO 

“1® 

+ 

M- 

O 

O 

41. 

1 1 o _ 

2 1 * 

f- 88. 

18. 

10- 

!_ 7 
* 8 - 

30. 

37*^15. 

42. 

3fs-' 

-96. 

19. 

15- 

L. 2 

' 3* 

31. 

45-3f 

43. 

18* 

+ta¬ 

20. 

5 

-8 * 

4. 

32. 

16$ -i- 10. 

44. 

3 6 • 
8"5 * 

il 

21 . 

TT 

-3. 

33. 

871 - 45. 

45. 

-w- 

51$» 


46 ‘ HI T6T- 49 - ft X yz ) X (1 -T-$). 

47 > (i+*) + (*“*). 50. tf-Kf-f). 

4 * (*-» + (¥ + *)• 


51. Toweling comes in 10-yd. bolts. How many towels 
30" long (finished length) could be made if each had a f" 
hem at each end ? 

52. For a certain garment 2f yd. of material is required. 
How many such garments can be cut from a 12-yd. bolt of 
the material ? 

53. A finished oak board 7f" wide is to be rip sawed into 
four strips of equal width. If the saw cut is y 1 /' in width, 
how wide will the strips be ? 

54. The sides of a low shed are 48" high. They 
are to be covered with a finished siding 7-§-" in 
width. These are overlapped as shown in the 
figure. How much must the lower six be over¬ 
lapped so that seven pieces will just cover the 
side of the shed ? 


I 

48 " 








74 


NEW HIGH SCHOOL ARITHMETIC 


55 . Two pieces of sheet iron 1 ft. wide are to be riveted 
together. Rivets are to be placed in holes with centers f" 
from the outside edges. Between these 
two rivets, five other rivets are to be T 
placed. What will be the distance be- \ 
tween the centers of two successive rivets ? *- 


I 


56 . How many yards of cloth at 66|^ per yd. can be 
bought for $ 8.00 ? 

57 . If 121 cords of wood cost $116.66 how much does one 
cord cost ? 

Complex Fractions 

52. A Complex Fraction is a fraction in which either the 
numerator or the denominator, or both, involve fractions. 

5 n 

Thus and — are complex fractions 1 . 

V * 

Since a fraction is merely an indicated division, complex 
fractions are simplified by performing the division accord¬ 
ing to the rule in § 51. 


Example 1. 

Simplify JL. 



9 

3 

Solution. 

f 5 . 10 _ ? 

x£ = 


■V 8 ' 9 0 

n 


2 

2 

Example 2. 

Simplify ?i. 

W 

ii 

Solution. 

2| _ 7 . 28 _ 7 

x«=: 


II 3 ’ 33 3 

ft 


EXERCISE 39 


i. if. 

3 . A. 

5 . ?k. 

7 . M. 

2 1 

To 

A 

2 

T 

A 

2 . A. 

a 

4 It 

* TT* 

6 . ?I. 

1 

8 . 5!. 

it 






FRACTIONS 75 



34 


3 9 


114 


6 6 


9. 

2 

14. 

8 

19. 

24. 

TTS 



H 


52’ 

6 A- 

2«’ 


10 . 

5f 

15. 

56 

20 . 

M 

25. 

1 + 

1 1 

TT 




38* 



¥ + 

26 * 
5 ~ 

11 . 

n 

16. 


21 . 


26. 

5|- 



2f 


^2T 





12 . 

1<H 

17. 

51 

22 . 

32 

o 

27. 

2*+lf 


2i' 



80 

T47 

8- 

54 

13. 

2f 

18. 

39 

23. 


28. 

3i + 2|. 


H' 


2 6 * 



A1 _ 
% 

i-i 


Fractional Parts 

53. Finding What Part One Number is of Another. 

Example. What part of 35 yr. is 10 yr. ? 

Solution. 1. 1 yr. is of 35 yr. 

2. Then 10 yr. is 10 x -jV or of 35 yr. 

3. .•. the result is or f. 

In this example, 10 yr. is regarded as a part of 35 yr., which is 
considered the whole. 

Rule. — To find what part one number is of another: 

1. Form the fraction whose numerator is the number re¬ 
garded as the part and whose denominator is considered the 
whole. 

2 . Reduce the fraction to lowest terms. 

EXERCISE 40 
1. What part of 60 is : 


a. 10 b. 15 c. 

20 

d. 25 

e. 35 

/. 45 

2 . What part of 108 is 
a. 6 b. 18 c. 

26 

d. 42 

CO 

CO 

^5 

/. 84 

3. 15 is what part of: 
a. 45 b. 100 c. 

80 

d. 135 

e. 210 

/. 350 






76 


NEW HIGH SCHOOL ARITHMETIC 


4. Determine what part 28 is of: 

a. 56 b. 72 c. 90 d, 133 e. 150 /. 140 

5. What part of Ilf is 5 ? is 14 ? is 20 ? is 10 ? 

6. What part of 24 is 2f? is 3f ? is 4f ? is 5f? 

7. What part of: 

a. 36 is 27 ? 5. 49 is 70? c. 2340 is 455 ? 

8. What part of: 

a - ii? if if ? c * It tt? 

9. 4f is what part of 3f ? 

10. What part of Ilf is 6 J ? 

11. What part of 40 is 6^- ? 

12. What part of 4^ is 7f ? 

13. What part of 2^ is 2^ ? 

14. 2765 is what part of 5690 ? 

15. 3f is what part of 20J ? 

16. What part of a long ton (2240 lb.) is a short ton 
(2000 lb.) ? 

17. Out of 260 gauze bandages made by a Red Cross 
group, 125 were made by one person. What part of the 
total was made by this person ? 

18. An auto tire, guaranteed to run 3500 miles, showed 
serious defects after being run 1050 miles. What part of 
the guaranteed milage was gotten ? 

19. In most states a bushel of corn on the cob weighs 
70 lb. and a bushel of shelled corn 56 lb. What part of a 
bushel of corn on the cob is a bushel of shelled corn? 

20. A cubic foot of pure ice weighs 56 lb. and of fresh 
water 62 J lb. What part of the weight of a cubic foot of 
fresh water is that of a cubic foot of pure ice ? 


FRACTIONS 


77 


21. A girl, employed at the rate of $ 5.00 per week, 

works days. How much should she receive ? 3 ' . ^ 

22. Out of 250 eggs placed in an incubator, 165 chicks 
were hatched. What part of the eggs placed in the incu¬ 
bator were lost ? 

If the eggs cost $ 12.50, what was the value of the eggs 
lost? 

23. A recipe for wild grape jelly calls for 1 pk. of wild 
grapes, 1 qt. of vinegar, 6 lb. of sugar, and i c. each of 
cloves and cinnamon. How much of each of these ingre¬ 
dients should be used with 4 qt. of grapes ? with 10 qt. ? 

24. A recipe for raspberry and currant preserves calls 
for 6 lb. of currants, 8 qt. of raspberries, and 6 lb. of sugar. 
How much sugar and raspberries should be used with 3 lb. 
of currants ? with 5 lb. ? with 3-i lb. ? 

25. A recipe for spiced currants calls for 7 lb. of currants, 
5 lb. of sugar, 3 T. of cinnamon, 3 T. of cloves, and 1 pt. of 
vinegar. 

How much of each of these ingredients should be used 
with 5 lb. of currants ? with 1 lb. ? with 13 lb. ? 

54. Finding a Number When Some Fractional Part of It is 
Given. 

Example 1. An auto takes 3 hours to go -f of a certain 
trip. How long will it take to go the whole distance at the 
same rate ? 

Solution 1. 3 hr. is § of the total time. 

2. .-. \ of the total time is £ of 3 hr. or 1^ hr. 

3. .*. | or the whole time is 5 X 1| hr. or 7| hr. 

Notice that is 3 \ or 3 x |. 

Example 2. 36 is f- of what number ? 

Solution. 1. } of the number is \ of 36 or 9. 

2. .•. the number is 7 x 9 or 63. 

Notice again that 63 = 36 -f-1 or 36 x 


78 


NEW HIGH SCHOOL ARITHMETIC 


Rule. — To find a number when a fractional part is given: 
Divide the number by the fractional part. 

EXERCISE 41 

1. 24 is f of what number ? Explain the solution. 

2. 35 is | of what number ? Explain the solution. 

3. 48 is a. -f of what number ? b. f of what number ? 

c. -§- of what number ? d. ^ of what number ? 

4. Of what number is 32 two thirds ? 

5. $ 500 is t 2 3 - of what amount ? 

6 . Of what number is 68 four fifths ? 

7. It took 72 days to complete f of the construction of 
a certain boat. How much longer will it take to complete 
the construction? 

8. A man received a bonus of $ 276 at the end of a year 
of work. If the bonus was f of his year’s salary, what was 
his regular salary ? 

9. A poultryman finds that he raises about as many 
chicks as he has eggs in his incubators. If he wants to 
have at least 500 chicks, how many eggs must he place in 
his incubators ? 

10. A carpenter finds that a given number of square feet 
of flooring will cover only about £ as many square feet of 
floor. How many square feet of flooring must he order so 
that he will have enough to cover 850 sq. ft. of floor surface ? 

Aliquot Parts 

55. An Aliquot Part of a number is a part of the number 
contained in it an integral number of times. 

Thus, 2 is an aliquot part of 10; also 2\, etc. 

56. The aliquot parts of 100 are particularly important. 

Complete the following: 


FRACTIONS 


79 


Table of Aliquot Parts of 100 


1 _ 

2 “ 

£= 

1 

8 — 

A — 

1 

16 

1 

3 ~ 

1 = 

1 = 

4 _ 

5 — 

3 _ 

8 “ 

5 _ 

8 — 

A = 

A = 

i 

H §k ! 
1 II 

1 _ 

4 ~ 

3 _ 

1 _ 

6 ~ 

5 _ 

7 _ 

8 — 

1 

it — 
i 

II II 

is-is 

¥ ~ 

¥ — 

9 ~ 

12 — 

i _ 

1 _ 

1 _ 

1 _ 

1 

TTF — 

5 “ 

7 

10 “ 

15 

A = 

Note. 

— Pupils should 

memorize 

as many of these as possible; 

they shoidd by all means 

memorize 

those printed 

in heavy black 


type. 

57. By means of the aliquot parts printed in heavy type, 
others can be determined. 

Thus, since ^ of 100 = 4, = 28. 

Again, since y 1 ^ of 100 is 6f, T \ is 26f. 

EXERCISE 42 

Give the following parts of 100: 


1 . 

f* 

8 . 

5 

T2* 

15. 

A* 

22 . 

4 

9 * 


29. 

A 

36. 

1 0 
11* 

2 . 

7* 

9. 

2 

TP 

16. 

A- 

23. 

A 


30. 

A- 

37. 

1 7 
25* 

3. 

2 

¥• 

10 . 

4 

1 1* 

17. 

A- 

24. 

A* 


31. 

A* 

38. 

11 
16* 

4. 

5 

¥* 

11 . 

3 

TP 

18. 

r 9 

TP 

25. 

A 


32. 

13 

2 5* 

39. 

1 3 
40* 

5. 

¥* 

12 . 

A* 

19. 

W* 

26. 

A* 


33. 

6 

16* 

40. 

1A 

1 6* 

6 . 

A* 

13. 

A 

20 . 

A 

27. 

A 


34. 

9 

1 1* 

41. 

1 1 
3TF* 

7. 

_7_ 

1 P 

14. 

TP 

21 . 

A- 

28. 

11 
12* 


35. 

1 1 
TP 

42. 

¥P 

What part of 

$ 1.00 is 









43. 

40 h 


47. 

60^? 

51. 


i 

9 

55. 

621 


44. 

70 


48. 



52. 

6i 

* 

9 

56. 

M' 

00 


45. 

4^? 


49. 

16f 


53. 

37 

1 

2 ' 

A 

57. 

871 

<h 

46. 

75 


50. 

33i 

0? 

54. 

£2 

t 

? 

58. 



59. 2-J-^? 60. 


80 


NEW HIGH SCHOOL ARITHMETIC 


58. The aliquot parts of $1.00 are used constantly in 
finding the cost of articles. 

Example 1. What is the cost of 75 lb. of sugar at 8-j^ 
per pound ? 

Solution. 1. = $ X V 

2. 75 lb. cost 75 X $ x h, or $ 6£ or $ 6.25. 

Example 2. What is the cost of 12| yd. of cloth at 
$ 1.75 per yard ? 

Solution. 1. Since 12 x X 175 — 175 X 12 x , 

the cost of 12f yd. at $ 1.75 per yard is the same as 
the cost of 175 yd. at 12| f per yard, or at $ } per yard. 

2. the cost is 175 x $ or $21| or $21,871. 

EXERCISE 43 

Find the cost: 

1. At 33per yard of : 

a. 15 yd. b. 24 yd. c. 12 yd. d. 18 yd. e. 20 yd. 
/. 10 yd. g. 8 yd. h. 11 yd. i. 5 yd. j. 16 yd. 

2. At 12-J- $ per yard of the quantities in Example 1. 

3. At per yard of the quantities in Example 1. 

4. At 81 f per yard of the quantities in Example 1. 

5. At 16|^ per yard of the quantities in Example 1. 

6. At 371 ^ per yard of the quantities in Example 1. 

7. At 75^ per yard of the quantities in Example 1. 

8. At 60^ per yard of the quantities in Example 1. 

9. At 87 |-1 per yard of the quantities in Example 1. 

10 . At $ 1.33i per yard of the quantities in Example 1. 

11 . At $ 1.40 per yard of the quantities in Example 1. 

12 . At $ 1.37^ per yard of the quantities in Example 1. 


FRACTIONS 


81 


Find the cost of: 
13. 30 articles at: 


a. 

33^-/ each 

b. 

40 fi 

each 

c. 

12i/ each 

d. 

$ 1.66| each 

e. 

16 if 

' each 

/• 

371/ each 

14. 

Of 240 articles 

at 





a. 

66f / each 

b. 

75? 

each 

c. 

6-j/ each 

d. 

$ 1.25 each 

e. 


each 

/• 

62i/ each 


15. Of 320 articles at the prices quoted in Example 13. 


16. Of 1200 articles at the prices quoted in Example 14. 


17. Of 25 articles at: 

a. 84/ each b. 73/each c. 12/each d. 19/each 


18. Of 50 lb. of articles of food at: 
a. 9/ per pound b. 17/ per pound 
d. 34 / per pound e. 47 / per pound 


c. 29 / per pound 
/. 69 / per pound 


19. Of 121 yd. of cloth at: 

a. 18 / per yard b. 24 / per yard 

d. 45 / per yard e. 69 / per yard 


c. 27 / per yard 
/. 98 / per yard 


20. Of 8^- doz. articles at: 
a. 36/ per dozen 
c. 84 / per dozen 
e. $ 1.20 per dozen 


b. 75/per dozen 
d. $ 1.00 per dozen 
/. $ 1.50 per dozen 


Find the total cost of the following purchases: 


21 . 15 lb. sugar at 81 / per pound. 
121 lb. prunes at 14 / per pound. 
12 cans of corn at 16f / per can. 
25 lb. oatmeal at 7 / per pound. 

22 . 35 lb. of oats at 21 / per pound. 
25 lb. barley at 31 / per pound. 


82 


NEW HIGH SCHOOL ARITHMETIC 


23. 50 lb. of coffee at 29^ per pound. 

25 lb. of coffee at 37 ^ per pound. 

15 lb. tea at 62-J- $ per pound. 

75 lb. tea at 37^ per pound. 

24. 17 tons coal at $ 12.50 per ton. ($ 12.50 = ^ of $100.) 

8 tons coal at $ 6J per ton. 

25. 150 bu. barley at $ 1.121 per bushel. 

250 bu. oats at 73 $ per bushel. (250 = i of 1000.) 

300 bu. wheat at $ 2.16f per bushel. 

Miscellaneous Fractional Examples 

EXERCISE 44 

Some of the following examples are of the nature of puzzles 
rather than practical. They are included because examples like 
them sometimes appear on examination papers, and because, 
being unusual, they will challenge the ability of the pupil. 

1. Find the sum of \\ of $5.67 and f of $8.75. 

2. If 13£ yards of cloth can be bought for $ 33.50, what 
will 15f yards cost at the same rate ? 

3. A, B, and C together received $17.40 for doing a 
piece of work. If A did ^ of the work, B and C the 
remainder, what should each receive ? 

4. Which is the greater, f of $ 16.12 or f of $ 17.67, and 
how much ? 

5. If 15f yards of a certain cloth can be bought for 
$12.73, how many yards of the same cloth can be bought 
for $18.76 ? 

6. Three men, A, B, and C, can do a piece of work in 18, 
24, and 36 hours respectively. How long will it take them 
to do the work if they work together? If they receive 
$11.25 for doing the work, how much should each receive? 

Suggestion. A does T V in one hour, B 2 y, and C 


FRACTIONS 


83 


7. A man left ^ of his estate to his wife, f pf the re¬ 
mainder to his son, and the rest to his daughter. The 
wife received $ 546.75 more than the daughter. "What did 
each receive ? 

8. A man spent f of his money for provisions, -J of the 
remainder for clothing, and had $ 12.87 left. How much 
had he at first ? 

9. If 5| pounds of coffee can be bought for $ 2.361, how 
many pounds can be bought for $ 6.38 ? 

10. A, B, and C can do a piece of work in 6, 14, and 21 
days, respectively. B and C worked alone for 5 days, when 
they were joined by A, after which the work was completed 
by them working together. If they received $216 for 
doing the work, how much should each receive ? 

11. A man left 1 of his property to his wife, \ to his 
elder son, to his younger son, 1 to his daughter, and the 
balance to a charitable institution. If the charitable insti¬ 
tution received $ 3697.50, how much did each of the family 
receive ? 

12. A man invested f of his wealth in real estate, of 
the remainder in railway shares, and the balance in city 
bonds. The amount invested in real estate exceeded by 
$ 8700 the amount invested in city bonds. Find the amount 
of each investment. 

13. A man divided $ 220 between his two sons in such 
a way that the younger received as much as the elder. 
How much did each receive ? 

14. An acre of ground contains 43,560 square feet. 

a. What part of an acre is a lot containing 5400 square 
feet? 

b. What part of an acre is a lot containing 7200 square 
feet ? 


84 


NEW HIGH SCHOOL ARITHMETIC 


c. If one of the smaller lots sold for $ 900, what would 
be the value of an acre at the same rate ? 

d. If one of the larger lots sold for $ 1050, what would 
be the value of an acre at the same rate ? 

15. A tank has two pipes. One of them can fill it in 
81 minutes, and the other can empty it in 12} minutes. 
How many minutes will it take to fill the tank if both are 
opened at the same time ? 


III. DECIMAL FRACTIONS 


59. Powers of Ten. — The numbers 100, 1000, etc., can 
be obtained by multiplying together a certain number of 
tens. 

Thus, 100 = 10 x 10, and 1000 = 10 x 10 x 10. 

Such numbers are called Powers of Ten. Notice that the 
number of zeros in the number is the same as the number 
of tens that must be multiplied together to give that 
number. 

Thus, 10,000 is the product of four tens. 

60. Decimal Fractions are fractions having as denomina¬ 
tors powers of ten, when written as follows : 

^ is written .3; T -§-g- is written .08 ; etc. 

The period placed before the number is called the Decimal 
Point. The figures to the right of the decimal point are 
said to be in the first decimal place , the second , the third , and 
so on. 

61. A figure written in the 

a. first decimal place denotes tenths; 

b. second decimal place denotes hundredths ; 

c. third decimal place denotes thousandths ; 

d. fourth decimal place denotes ten thousandths; 

e. fifth decimal place denotes hundred thousandths ; 

/. sixth decimal place denotes millionths ; etc. 

Thus, .037 means + T fb- + T ^W- 

But xihr + ioVfr — Tinhr + nrsrff = 

and .•. .037 = 


85 


86 


NEW HIGH SCHOOL ARITHMETIC 


Similarly, .245 means T 2 S + 

But x 2 o + t^o + raW = twit + T&ita + too = iW?f> 
and .-. .245 = tffo. 

Rule. — To read a fraction written in decimal form: 

1. Read the number on the right of the decimal point as if it 
were an integer, disregarding any zeros immediately to the right 
of the decimal point. 

2. Then repeat the name of the fractional part indicated by the 
right-most digit. 

3. If there are digits to the left of the decimal point, read the 
number indicated by them, then the word “ and,” and finally the 
decimal fraction. 

Thus, .261 is read two hundred sixty-one thousandths; 

.0261 is read two hundred sixty-one ten thousandths; 

.000261 is read two hundred sixty-one millionths; etc. 

Also, 2.305 is read two and three hundred Jive thousandths. 


EXERCISE 45 

1. What fractional part is indicated by a number in the 
third decimal place ? in the fifth ? in the second ? in the 
seventh ? in the eighth ? 

Read the following numbers: 


2 . 

.53. 

6 . 12.052. 

10 . 24.0071. 

14 . 

2574.009. 

3 . 

.183. 

7 . 90.065. 

11 . .84073. 

15 . 

67.0203. 

4 . 

.995. 

8 . .0039. 

12 . 3.689313. 

16 . 

736.89245, 

5 . 

4.27. 

9 . 8.208. 

13 . 5.90854. 



17 . 

a. Compare the fractions .20 and .2. 




b. Compare the fractions .32 and .320. 

c. Compare the fractions .5, .50, and .500. 

d. What effect upon the value of a decimal fraction 
does it have to annex zeros to the right of it? 

18 . What is the common value of .4, .40, .400, etc. ? 


DECIMAL FRACTIONS 


87 


19 . a. Determine the value of .2 and .02. (2 is called 

the significant figure.) 

b. Determine the values of .3, and .003. 

c. Determine the values of .5 and .0005. 

d. Does it change the value of a decimal fraction to 
place one or more zeros between the decimal point and the 
first significant figure ? 

20. What is the effect upon a decimal fraction of placing 
between the decimal point and the first significant figure: 

a. one zero; b. two zeros; c. three zeros ? 

Write the following numbers : 

21. Fifty-two and forty-three hundredths. 

22. One hundred fifty-eight thousandths. 

23 . Nine and thirteen ten thousandths. 

24 . Fifty-nine thousand, three hundred ninety-eight 
millionths. 

25. Twenty-six and eight hundred five thousand, three 
hundred three millionths. 

26 . Read the following sentences based upon statistics 
in a recent Bulletin of the U. S. Bureau of Labor. 

The average wholesale price during the year 1915: 

a. Of lath was $ 3.8386 per thousand. 

b. Of single-strength window glass in small sizes was 
$2.4225 per 50 square feet. 

c. Of tin plate was $ 3.2417 per 100 lb. box. 

d. Of barbed wire fencing was $ 2.5348 per 100 lb. 

27. If a bar of steel at a tefnperature of 32° F. is heated 
lJ-°, its length will increase .000011 of itself, a bar of iron 
.000012 of itself, a bar of copper .000017 of itself, and a 
bar of lead .000027 of itself. (From a textbook on 
Physics.) 


88 


NEW HIGH SCHOOL ARITHMETIC 


28 . The diameter of No. 1 wire is given as follows on 


different measuring instruments: 

Brown and Sharpe instrument.2893 in. 

Washburn instrument.2830 in. 

U. S. Standard.2812 in. 

(From a textbook for machinists.) 


62. Changing a Decimal Fraction to a Common Fraction. 

Example 1. 11.28 = 11 + ywo ~ W’ 

Example 2. .0523 = iwr o- 

Example 3. 2.035 = 2 + yffo == fUf. 

Rule. — To change a decimal fraction to a common fraction : 

Write a fraction having for its numerator, the given digits 
with the decimal point omitted; and for its denominator, 1 
followed by as many zeros as there are places to the right of 
the decimal point. 

Thus, 2.065 = 5 32.05 = ; and .0347 = rff** 


EXERCISE 46 

Express the following as common fractions in their low¬ 
est terms: 


1 . 2 . 8 . 

2. .005. 

3 . .684. 

4 . 1.85. 

5 . 8.512. 

6. 75.44. 

7 . 30.75. 


8. .1975. 

9 . 68.461. 

10. .025. 

11. .0376. 

12. .0096. 

13 . 3.0875. 


14 . .08309. 

15 . .00125. 

16 . .0625. 

17 . 8.75. 

18 . 25.25. 

19 . .0375. 


20. 4.235. 

21. 10.25. 

22. 83.125. 

23 . 6.25. 

24 . 13.042. 

25 . 5.375. 


63. Adding and Subtracting Decimals. 

When numbers involving decimals are written so that 
the decimal points are in a vertical line, the digits repre¬ 
senting the same fractional parts fall in vertical columns. 
Consequently, the numbers can be added at once. 





DECIMAL FRACTIONS 


89 


Example 1. Add 7.89, 31.4, and .086. 

Solution. The explanation of this solution is left 
to the class. 


If desired, zeros may be added to the right of some 
of the numbers in order to make all of them have the 
same number of decimal places, as arranged at the right. 

Example 2. Subtract 98.725 from 162.07. 


7.89 

31.4 

.086 

39.376 

7.890 

31.400 

.086 

39.376 


Solution. The solution may be arranged as at the 
right. Notice that a zero is annexed to the right of the 
minuend in order to make it and the subtrahend each 
have three decimal places. 


162.070 

98.725 

63.345 


EXERCISE 47 

Add the following: 

1. 25.5, 2.0076, 1.7852, and 3.084. 

2. 2.601, .9763, 35.083, and .00745. 

3. 165, .94468, .0051, and 39.236. 

4 . 7.3075, 8.62, .03874, and 2.319. 

5 . 5.0805, .00007, 2.637, and .014976. 

6. .39665, 93.3896, .0562, and 805.74. 

7 . 9.9, .00616, 75.1208, and 16.004. 

8. 5.06212, 35.49, 87.65, 4290.24, and 297.638. 

9. 2.0342, 69.21, 8.463, .12058, and 3.028. 

10. 14.069, 20.743, .085, 9.263, and .00584. 


Subtract: 

11. .4167 from 5.2765. 

12. .0825 from .32943. 

13 . .317 from 2. 

14 . .00964 from .0205. 

15 . 2.0843 from 11.365. 


16 . .5431 from 2.9063. 

17 . .06397 from 10. 

18 . .08948 from 3.52866. 

19 . 47.6007 from 830.453. 

20. .00099 from .06009. 





90 


NEW HIGH SCHOOL ARITHMETIC 


Find the sums: 


21 . 

22 . 

23 . 

24 . 

25 . 

2516.34 

809.142 

1702.3 

4617.335 

33.862 

375.081 

76.053 

56.7798 

11.008 

518.0325 

9.004 

4820.279 

231.0205 

700.8318 

7.664 

61.2865 

36.0086 

7.88 

4.215 

43.2218 

649.75 

7.251 

339.021 

63.08 

.9086 

31.1092 

237.85 

65.7081 

3010.508 

8201.22 

3750.546 

49.9957 

.9959 

715.244 

66.192 

879.0732 

3654.0075 

8.114 

29.9592 

735.11 

.5531 

68.132 

563.7732 

2.007 

5.368 

25.864 

911.0768 

8104.22 

687.14 

71.6842 


64. Multiplying Decimals. 

Example 1. .3x.2=^x = yfg- = .06. 

Example 2. .4 x .12 = ^ x = yfjy = -048. 

Example 3. .25 x .83 = ^ x ^ = - 2075 - 

Notice that the number of decimal places in the product 
in each of these examples equals the sum of the number in 
the multiplier and in the multiplicand. Hence, 

Rule. — To multiply a decimal by a decimal: 

1. Multiply the numbers, disregarding any decimal points. 

2 . Point off from the right as many decimal places as there 
are in the multiplier and the multiplicand together, prefixing 
zeros to the product if necessary. 


Example 4. Multiply .764 by .0108. 

Solution. In this example, the sum of the number .764 

of decimal places in the multiplier and in the multi- .0108 

plicand is seven. After multiplying, there are only 6112 

five figures in the product. Two zeros are therefore 7640 

prefixed to the product 82512. .0082512 










DECIMAL FRACTIONS 


91 


EXERCISE 48 

Multiply: 


1 . 

8.27 by 27.3. 

13 . 

.0782 by 63.05. 

2 . 

.096 by 5.61. 

14 . 

.9256 by .0821. 

3 . 

.0708 by .365. 

15 . 

35.702 by .012. 

4 . 

.6528 by 9.7. 

16 . 

42.004 by 3.404. 

5 . 

5.648 by .082. 

17 . 

.2705 by .234. 

6 . 

2.835 by 34.1. 

18 . 

9.631 by 3.1416. 

7 . 

45.76 by .0207. 

19 . 

2.8 by .7854. 

8 . 

.5114 by .3043. 

20 . 

3.8 by .00011. 

9 . 

.0682 by 16. 

21 . 

1.85 by 26.3 by 4.32. 

10 . 

4.486 by 5.38. 

22 . 

5.06 by 3.14 by .243. 

11 . 

18.052 by .75. 

23 . 

65.842 by .85 by .125. 

12 . 

21.96 by 4.78. 

24 . 

.875 by 2500 by .05. 

25 . 

At 39° F. temperature (approximately), a cubic foot 


of copper weighs 8.45 times as much as water ; of cork, .24 
times as much as water; of gold, 19.26 times as much; of 
ice, .92 times as much; of lead, 11.35 times as much; of 
steel, 7.83 times as much; of air, .001293 times as much. 

Determine the weight of a cubic foot of each of these 
materials if a cubic foot of water weighs 62.5 pounds. 

26 . If an object is allowed to fall freely from any height 
for any number of seconds, the distance through which it 
will fall may be found by multiplying 16.08 feet by the 
square of the number of seconds. (The square of a number 
i3 that number multiplied by itself.) 

Determine how far an object will fall in: 

a. 3 seconds ; b. 4 seconds; c. 5 seconds. 

Determine also how far it falls during the 4th second, and 
during the 5th second. 


92 


NEW HIGH SCHOOL ARITHMETIC 


27 . A farmer sold 15 loads of wheat, averaging 83.25 
"bushels to a load, at $1.95 per bushel. How much money 
was due him ? 

28 . Eind the cost of 12 bales of silk averaging 73.625 
yards to a bale, at $ 1.93 per yard. 

29 . The width of the slot, for the screw driver, in the 
head of a flat-head machine screw, is found by multiplying 
the diameter of the body of the screw by .173 and then 
adding .015. 

Thus, if the diameter is .086 inch, the width of the slot is 
(.173 x .086 + .015) inch. 

a. Compute the result for this illustration. 

b. Compute the width of the slot when the diameter is 
.268 inch. 


65. Multiplying by 10, 100, 1000, etc. 


Example 1. 


10 x .3 = 10 x t 3 o = 3. 


10 x .03 = 10 x 
10 x .003 = 10 x 


3 _ 3 _ Q 

TOO — yq- — .o. 

3 _ 3 _ HQ 

TOOT — TTRT — • U,x 


Example 2. 100 x .3 = 100 x = 30. 

100 x .03 = 100 x yfo = 3. 

100 x .003 = 100 x yo 3 ^ = * = .3. 

Multiplying a decimal by 10 has the effect of moving the 
decimal point of the multiplicand one place to the right; 
multiplying by 100, of moving the decimal point two places 
to the right. Hence, 


Rule. — To multiply a number by 10, 100, 1000, etc., move 
the decimal point of the multiplicand as many places to the 
right as there are zeros in the multiplier, annexing zeros to the 
multiplicand if necessary. 

Thus, 100 x 2.653 = 265.3; 

1000 x 3.84 = 3840. 



DECIMAL FRACTIONS 


93 


66 . Multiplying by .1, .01, .001, etc. 


Example 1. .1 x .3 = .03. 


.1 x .03 = .003. 
.1 x .003 = .0003. 


Example 2. .01 x .3 = .003. 


.01 x .03 = .0003. 
.01 x .003 = .00003. 


Multiplying a decimal by .1 has the effect of moving the 
decimal point of the multiplicand one place to the left; 
multiplying by .01, of moving the decimal place two places 
to the left. Hence, 

Rule. —To multiply a number by .1, .01, .001, etc., move the 
decimal point of the multiplicand as many places to the left as 
there are decimal places in the multiplier, prefixing zeros to the 
multiplicand if necessary. 

Thus, .001 x 5 = .005; and .0001 x 26.3 = .00263. 


EXERCISE 49 


1. Multiply each of the following numbers by 10. 

a. 27.6 b. 3.85 c. 2.608 d. .5703 e. 642 
/. 30.451 g. 51.008 h. 431.8 i. 500.09 j. .089 

2. Multiply each of the numbers in Example 1 by 100. 

3 . Multiply each of the numbers in Example 1 by 1000. 

4 . Multiply each of the numbers in Example 1 by .1. 

5. Multiply each of the numbers in Example 1 by .01. 

6. Multiply each of the numbers in Example 1 by .001. 
Multiply: 

7. 85.2 by 10. 12. 5839 by .00001. 

8 . 377 by .01. 13 . 1.417 by 10,000. 


9 . .0065 by 1000. 

10. .0208 by 100. 

11. 12.61 by .01. 


14 . 368.8 by .001. 

15 . 937,628 by .00001. 

16 . .06023 by 100,000. 


94 


NEW HIGH SCHOOL ARITHMETIC 


What is the cost of: 

17 . 10 dozen eggs at $.25 per dozen? 

18 . 100 feet of oak lumber at 12.5 ^ per foot ? 

19 . 1000 bushels of oats at 93.75^ per bushel? 

20. 100 feet of garden hose at 13.25 ^ per foot ? 

21. 1000 tons of chestnut coal at $ 5.3256 per ton ? 

Note.— In this and the following examples, the prices aro 

average wholesale prices. 

22. 1000 tons of anthracite egg coal at $ 5.0446 ? 

23 . 1000 pounds of Ohio wool at $.5714 per pound? 

24 . 1000 pounds of worsted yarn at $.7875 per pound? 

25 . 100 gallons of denatured alchohol at $ .3717 per 
gallon ? 

26 . 100 pounds of candles at $ .0725 per pound ? 

27 . 1000 yards of all wool French serge at $.3325 per 
yard ? 

28 . 10,000 yards of all wool storm serge dress goods 
at $ .5574 per yard ? 

29 . 100 dozen men’s cotton shirts at $4.1875 per dozen? 

30 . 100 yards of Middlesex suiting, 15 ounce variety, 
55-56 inches wide, at $ 1.5638 per yard ? 

31 . 1000 yards bleached muslin shirting at $.0782 per 
yard? 1000 yards bleached muslin shirting at $.1001 per 
yard? 

32 . 10,000 pounds of linen shoe thread at $ 1.0763 per 

pound ? 1000 spools of cotton thread at $ .0359 per spool ? 

33 . 100 barrels of salt pork at $ 18.3173 per barrel ? 

34 . 10,000 quarts of milk at $ .0376 per quart? 

35 . 100,000 pounds of poultry at $ .1373 per pound ? 

36 . 10,000 pounds of fresh beef at $.1255 per pound ? 

37 . 100,000 dozen eggs at $ .2657 per dozen ? 

38 . 10,000 gallons of denatured alcohol at $ .3717 per 
gallon ? 


DECIMAL FRACTIONS 


95 


Division of Decimals 


67. Dividing a Decimal by an Integer. 

Example 1. .8 -f- 2 = 2 = = .4. 

Example 2. .045 -- 3 = ■+■ 3 = yjfo = .015. 

Example 3. 12.18 -- 6 = i- 2 ^ -f- 6 = f= 2.03. 

Notice that the number of decimal places in the quotient 
in each example is the same as the number in the dividend. 
Hence, 


Rule. — To divide a decimal by an integer : 

1. Carry out the division, disregarding the decimal point. 

2 . Point off from the right in the quotient as many decimal 
places as there are in the dividend. 

The following example illustrates the most convenient 
arrangement of the solution. 


Example 4. Divide 4742.66 by 754. 

Solution. Notice that each figure of the quo¬ 
tient is directly above the last figure of the partial 
product obtained by multiplying the divisor by that 
figure of the quotient. Notice also that the decimal 
point of the quotient is directly above that of the 
dividend. 


6.29 

754 [4742.66 
4524 
218 6 
150 8 
67 86 
67 86 


Example 5. Divide 237.5 by 68, carrying out the divi¬ 
sion to three decimal places. 

3.492 
68 1237.500 
204 


Solution. Notice that two zeros are annexed to 33 5 

the dividend so that there will be three decimal 27 2 

places in it. 6 30 

The final result may be written 3.492+ 0 12 

180 










96 


NEW HIGH SCHOOL ARITHMETIC 


68 . Dividing a Number by a Decimal Divisor. 


Notice the following divisions. 

4 4 

2j8 ; also 20[80 ; also 


4 


Observe that the dividend and the divisor in the second 
division have each been multiplied by 10, and in the third 
division have each been multiplied by 100. All three 
quotients, however, are the same. Hence, 

Rule I. — The dividend and the divisor may each be multi¬ 
plied by 10, 100, etc., without changing the value of the quo¬ 
tient. 

This fact aids in solving 

Example 1. Divide 6.804 by 2.1. 

Solution. The divisor can be made an integer by multiplying 
it by 10. Hence multiply both dividend and divisor by 10. 

Then the solution of ]. 

SJjftiOi the same as 

The second dividend and divisor are 
obtained by moving the decimal points 
of the original ones one place to the 
right. 

Rule II. — To divide a number by a decimal divisor: 

1. Move the decimal point of the divisor to the right until 
the divisor becomes an integer; then move the decimal point of 
the dividend an equal number of places to the right, annexing 
zeros as may be necessary. (See Rule above.) 

2 . Then carry out the division as directed by the Rule in 
§ 67. 


the solution of 
3.24 
2lf6a04 
63 
50 
42 
84 
84 




DECIMAL FRACTIONS 


97 


Example 2. Divide 2.75918 by .7261. 3.8 

Solution. Notice that the original decimal ?7261.[ 2?7591.8 
points have been crossed out, and new ones in 2 1783 

serted four places to the right. 5808 8 

5808 8 

Example 3. Divide .06084 by 2.34. 


Solution. In this example, there must be three 
decimal places in the quotient, of which 2 and 6 
occupy the second and third places. A zero must 
be prefixed to 26 before locating the decimal point, 
in order to have as many places in the quotient as 
in the dividend. 


.026 

2*34.|?06.084 
4 68 
1404 
1404 


EXERCISE 50 


Divide. If the division is 

not exact, obtain a quotient 

having three decimal places. 


1. 15 by .3. 

13 . 46.35 by .15. 

2 . 28 by .04. 

14 . .48 by .0012. 

3 . 3.6 by .6. 

15 . 284.24 by .038. 

4 . .42 by .7. 

16 . 4.6292 by 2.84. 

5 . 5.4 by .03. 

17 . .49245 by .549. 

6 . .66 by 3. 

18 . 29,379.7 by .47. 

7 . 8.4 by 1.2. 

19 . 19.7635 by 8.41. 

8 . 9.6 by .32. 

20 . .31236 by .00685. 

9 . 1.24 by .004. 

21 . 462.90881 by .9107- 

10 . 25.6 by .008. 

22 . 691.8015 by 71.74. 

11. 100 by .05. 

23 . 609.4429 by .01234. 

12 . 376.8 by .006. 

24 . 332.45 by .0488. 

In the following problems, carry out the result as in- 

dicated. 


25 . During the year 1931,97,139 immigrants came to this 


country. What was the average number per month ? (One 
decimal place.) 









98 


NEW HIGH SCHOOL ARITHMETIC 


26 . How many miles will a train run in one hour at the 
rate of one mile : 

a. in V 20" (1 minute and 20 seconds) ; b. in 2' 3"; 
c. ini' 43"; d. ini' 10"? 

27 . If it takes a train 4 hours and 20 minutes to run 140 
miles, what is the rate per hour ? (One decimal place.) 

28 . A short ton of coal*occupies about 38.4 cubic feet of 
space. About how many tons of coal can be placed in a 
bin containing TOO cubic feet? (Two decimal places.) 

29 . On an auto trip of 157 miles, 14 gallons of gasoline 
were consumed. How many miles did the machine run on 
one gallon of gasoline ? 

30 . An autoist finds that his four tires are practically 
worn out at the end of 10,250 miles of service. If the tires 
cost him $23.75 each, what was the expense for tires in 
cents per mile? (Cents and one decimal.) 

31 . From 9,345,000 acres of land in Illinois, 238,298,000 
bushels of corn were gathered in 1930. How many bushels 
per acre were harvested, correct to tenths ? 

32 . In Iowa, 360,750,000 bu. of corn were harvested from 
11,100,000 acres. What was the average number of bushels 
per acre, correct to tenths ? 

33 . In 1930, the population of the United States was 
122,775,046. The total land area was 2,973,890 sq. mi. 
What was the average number of people per square mile in 
the United States in 1930? 

34 . In 1928, there were 25,179,696 children enrolled in 
the schools of the country. The total cost of instruction 
was $2,180,558,660. What was the cost per pupil, correct 
to tenths of a cent? 

35 . In a recent year, there were in the United States 
24,825 telegraph offices. 


DECIMAL FRACTIONS 


99 


a. From these offices, 75,135,405 messages were sent. What 
was the average number per office ? (One decimal place.) 

b. The net profits of the operating companies were 
$ 7,274,900. What was the net profit per office ? (One 
decimal place beyond cents.) 

36 . The exports of certain articles of food from the 
United States during the fiscal year ending June 30 in the 
last year before the Great European War and in the year 
1916, the second year of that war, were: 


Article 

The Year Ending June 30, 
1914 

The Year Ending June 30, 
1916 

Fresh beef 

16,394,404 

1231,215,075 

Bacon 

193,964,252 

579,808,786 

Fresh pork 

2 ,668,020 

63,005,524 

Butter 

3,693,597 

13,503,279 

Cheese 

2,427,577 

44,394,251 


How many times as great was the value of the exports 
in 1916 as they were in 1914 ? (One decimal place.) 


69. Dividing a Decimal by 10, 100, 1000, etc. 

By § 67, 84.6 10 = 8.46. 

Similarly, 92.4 - 100 = .924. 

Rule. — To divide a number by 10, 100, 1000, etc. 

Move the decimal point of the dividend as many places to 
the left as there are zeros in the divisor. 

An important application of this rule is found in 


Example 1. Divide 4716.28 by 62,800. .07 51 

Solution. In order to remove the two zeros 628.00? 1 47.36^28 
from the divisor, move the decimal point two 43 96 

places to the left, by dividing by 100. Then 3 20 2 

also move the decimal point of the dividend 3 14 0 

two places to the left . 6 28 

This does not change the quotient. 6 28 













100 NEW HIGH SCHOOL ARITHMETIC 
EXERCISE 51 

1. Divide each of the following numbers by 10. 

a. 37.8 b. 92.83 c. 5.637 d. 426.54 e. .695 

/. 10.342 g. .008 h. .125 i. 60.086 j. 402.54 

2. Divide each of the numbers in Example 1 by 100. 

3 . Divide each of the numbers in Example 1 by 1000. 

4 . Divide each of the numbers in Example 1 by 200. 

5 . Divide each of the numbers in Example 1 by 500. 


Divide: 


6 . 

630 by 500. 

12 . 

5156.8 by 4000. 

7 . 

2697.5 by 8300. 

13 . 

27,865 by 3000. 

8 . 

683.37 by 270. 

14 . 

82,648 by 2400. 

9 . 

.463 by 1000. 

15 . 

2,756,830 by 4300. 

10 . 

7.815 by 2000. 

16 . 

1,276,534 by 478,000. 

11 . 

85,536 by 648,000. 

17 . 

2,376,583.45 by 27,500, 

18 . 

How many tons of coal 

are there in 7680 pounds ? 


What is the value at $8.75 per ton? 

19 . How many thousand feet of lumber are there in a 
pile of wood containing 15,350 feet ? What is the value at 
$ 36 per thousand ? 

20. How many hundreds of pounds of glucose are there 
in three shipments consisting of 2346, 5485, and 4275 pounds 
respectively ? What is the value at the price $ 2.2942 per 
hundred ? 

21 . A “ square ” of roofing material is 100 square feet. 
How many squares of roofing material are required to cover 
a roof containing 2760 square feet ? 

22. How many tons of cabbage are there in three wagon 
loads consisting of 1675 pounds, 1845 pounds, and 1780 
pounds ? What is the value at the average price of $ 13.05 
per ton ? 


DECIMAL FRACTIONS 


101 


23 . How many hundreds of pounds of tobacco are there 
in five loads weighing 2765, 2430, 3125, 2875, 2945 pounds 
respectively ? What is the value of the whole lot at $3.78 
per 100 pounds? 

24 . Find the value of 1578 pounds of beans at $5.81 per 
hundred. 

25 . Find the value of three lots of sheep weighing 
respectively 1345 pounds, 2253 pounds, and 1682 pounds 
at $ 5.92 per hundred. 

26 . Find the value of five lots of steers weighing respec¬ 
tively 5765, 8342, 9724, 7586, and 8794 pounds at $11.35 
per hundred pounds. 

27 . Find the value at $ 15.72 per hundred of; five hogs 
weighing respectively 258, 274, 264, 278, and 285 pounds. 

28 . Find the value of 12,500 ft. of lumber at $34 per 
thousand. 

29 . Find the value of 6500 shingles at $5.75 per 
thousand. 

30 . Find the cost of 775 lb. of oil meal at $2.10 per 
hundred. 

31 . Find the cost of 1500 lb. of bran at $1.15 per hundred. 

32 . Find the cost of 650 lb. of middlings at $ 1.25 per 
hundred. 

33. Find the cost of 4500 pads of paper at $ 37.50 per 
thousand. 

34 . Find the cost of 1155 lb. of timothy seed at $ 8.35 
per hundred. 

35. Find the cost of 845 lb. of clover seed at $ 36.85 per 
hundred. 

36 . Find the cost at $ 2.15 per hundred of : 
a. 940 lb.; b , 1125 lb.; c. 650 lb. 


102 


NEW HIGH SCHOOL ARITHMETIC 


37 . Find the cost at $ 27.50 per thousand of : 

a. 4500 ft. of lumber; b. 3750 ft.; c. 8475 ft. 

38 . Find the cost at $ 8.25 per hundred of: 

a. 275 lb. of timothy seed; b. 169 lb.; c. 735 lb. 

39 . Find the cost at $ 37.50 per hundred of: 

a. 624 lb. of clover seed; b. 369 lb.; c. 472 lb. 

40. Find the value at $ 9.86 per hundred of: 

a. 12,400 lb. of pork; b. 8250 lb.; c. 7785 lb. 

70. Changing a Common Fraction to a Decimal Fraction. 

Example 1. Change T ^- to decimal form. 

Solution. 1. = 57- 125 = .456. .456 

125157.000 
50 0 
7 00 
6 25 
750 
750 

In some cases, there is not an exact decimal equivalent 
for a given common fraction. In such cases, a decimal is 
often derived which represents the given fraction for a 
certain number of decimal places. 

Example 2. Express as a decimal fraction correct 
to three decimal places. 

.583 

Solution. 1217.000 

60 
100 

96 T \ = .583+ 

40 

36 

4 

Sometimes the result is written .583 x \ or .583^; but this is 
seldom a useful form of the result. 




DECIMAL FRACTIONS 


103 


A more common practice is to increase the last figure of the 
decimal by 1 if the remainder is equal to or more than one half 
the divisor. 

Thus, .725f may be written .726, 
and .725| may be written .725. 


EXERCISE 52 

Change the following fractions to decimals, giving the 
first three decimal places: 


1. f 

5. 

21 m 

9 - a- 

CO 

H 

17. 

T6* 

2- i- 

6. 

1 

9* 


10- «• 

14. *. 

18. 

A* 

3- A- 

7. 

2 

T* 


ii. f 

15. if. 

19. 

11 
12* 

4. }. 

8. 

A 

• 

12. M* 

16. -V. 

20. 

23 

1 6* 

21- -W- 



26. 

At* 

31. 

35 7 
¥4¥* 


oo 2 5 6 

-yjp. 



27. 

1 9 

18 9* 

32. 

23 

2 T 


23 * AVo- 



28. 

1562 

7T * 

33. 

5 7 54 

2 36 * 


24. 



29. 

2 5 3 

34. 

8 2 75 




8100* 

94382* 


oc 4 33 



30. 

31 

35. 

2457 5 


25 * 12 80* 



430* 

123582* 


36. Change f to decimal form by 
of 100. 

c 7 . 8 3 j. 100 37| o 7 i _ 

SoluUou. - = - of — = —= .37 - 

use of “ aliquot parts 

= .375. 

Similarly express 

in decimal form 

• 



37. 

41. 

1 

¥* 


45. f. 

49. f. 

53. 

f* 

38. 

42. 

3 

z- 


46. T V 

50. 

54. 

4 

0* 

39 i. 

43. 

1 

6* 


47. T V- 

51 * T6* 

55. 

A* 

40. f. 

44. 

I- 


48. J. 

52. T V- 

56. 

A* 







104 


NEW HIGH SCHOOL ARITHMETIC 


Miscellaneous Examples 

EXERCISE 53 

1. Eind the weight of a cubic inch of material if a cubic 
foot weighs the amount indicated below, by dividing by 
1728, to three decimal places : 

a. cast iron, 450 lb. per cubic foot. 

b. steel, 489 lb. per cubic foot. 

c. lead, 711 lb. per cubic foot. 

2 . The number of cubic inches in a steel shaft 2 in. in 
diameter and 12 ft. long is 

.7854 x 12 x 12 x 2 x 2. 

Determine the number of cubic inches, and then from the 
result of part b of Example 1, find the weight of the shaft. 

3. a. Eind the total amount due a farmer who sold to a 
mill: 

1468 bu. of oats at 23 ^ per bushel. 

545 bu. of wheat at 54 $ per bushel. 

b. The farmer bought: 

800 lb. of bran at $ 1.15 per hundred pounds. 

200 lb. of oil meal at % 2.15 per hundred. 

400 lb. of middlings at $ 1.10 per hundred. 

c. How much did the mill man owe the farmer, or the 
farmer owe the mill man ? 

4. a. A gallon contains 231 cubic inches ; a cubic foot, 
1728 cubic inches. How many gallons of water will go 
into a vessel having a capacity of 1 cubic foot ? 

b. A cubic foot of water weighs about 62.5 pounds. 
What is the weight of a gallon of water, correct to two 
decimal places? 

5. Eind the total cost of : 

78 lb. of goods at $ .12^ per pound. 

54 lb. of goods at % .40 per pound. 


DECIMAL FRACTIONS 


105 


In Examples 6 to 13, find the result correct to one deci¬ 
mal place. 

6. The horse power developed by a certain steam engine 
is found by completing the following indicated computation: 

40 x 4 x 63.62 x 120 
3 x 33000 

7. The number of pounds of pull necessary to raise 
120 pounds of weight a distance of 30 inches by means of a 
certain combination of pulleys is found by completing the 
following indicated computation: 

120 x (4.5 - 3) 

2 x 4.5 

8 . To find the number of threads per inch on a standard 
machine screw, divide 6.5 by the diameter of the screw 
plus .02. 

Thus, if the diameter is .0828 in., the number of threads 
6.5 

18 .0828 + .02' 

a. Find the result in this example. 

b. Find the result also if the diameter is .1210 in. 

9. The length of a belt running around a 14-in. pulley 
and a 20-in. pulley, placed 12 ft. apart, is found by solving 
the following indicated computation: 

3.1416(10 + 7) + (2x 144) + 

10 . 1400.4 -3.28. 


11 . 


12 . 1 + 


13 . 


2116.3 x 12.39 
273 

96.056 


.168 X 1400.4 
5 x 86.878 x 333 


144 x 64.7 








106 


NEW HIGH SCHOOL ARITHMETIC 


MASTERY TEST I 


1. Add: 

2. 

Subtract: 

5038 


99807 

671 


46395 

20563 

9017 

5. 

Multiply: 

265 


3.1416 

8049 


.38 

672 

38 

8. 

Divide: 

2764 

2.43)65.7628 


3. Multiply: 4. Divide: 

3 v 16 v 6 71 • 3 

8 X T X Ti * 2 • -g- 


6. Divide: 


1 5 3 

T6 • 


7. Mdd: 

A + f + If 


9. Multiply: 

H x Si 


10 . 


Subtract: 

3| -2* 


11 . W T hat is the cost, at 48 $ per hundred pounds, of three 
loads of potatoes weighing 1785 lb., 1854 lb., and 1926 lb., 
respectively ? 

12. If 361,090,000 bu. of potatoes were raised on 
3,394,000 acres, what was the average number of bushels 
raised per acre ? 

MASTERY TEST II 


1. Add: 

2. 

Subtract: 

3. 

Multiply: 

4. Divide: 

72.48 


265.834 

9 

T6 

x -y>- x t 

15^3| 

365.30 


90.867 



2706.42 






853.95 

5. 

Multiply: 

6. 

Divide: 

7. Add : 

27.06 


763.08 


+ 

< 

+ 2 J + 41 

98.25 


.785 




360.04 

7059.38 

8. 

Divide: 

9. 

Multiply: 

10. Subtract: 

2604.76 

3.704)928.561 


2ix4f 

16| - 54 


11 . What is the cost, at $ 7.25 per ton, of four loads of 
coal weighing 4265 lb., 4370 lb., 4550 lb., and 4475 lb., 
respectively ? 

12 . If 50,234,000 bu. of rye were raised on 3,722,000 
acres, how many bushels were raised per acre ? 










DECIMAL FRACTIONS 


107 


PROBLEMS TEST 

1 . Prepare the extensions for and find the total of an in¬ 
voice containing the following items: 

275 yd. of silk @ $ 2.43 per yard. 

215 yd. of rayon @ 63 ^ per yard. 

142 yd. of serge @ $1.85 per yard. 

185 yd. of prints @ 32 ^ per yard. 


2 . 


The Population op 

In 1920, Was 

In 1930, Was 

California .... 

3,426,861 

5,677,251 

Georgia .... 

2,895,832 

2,908,506 

Maine. 

768,014 

797,423 

Ohio. 

5,759,394 

6,646,697 


а. Find the increase in the population of each state dur¬ 
ing the 10-year period. 

б. Find the average increase per year during the period, 
for each state. 

3. At $ 12.35 per hundred pounds, what is the value of 
five head of cattle which average 1372 lb. each? 

4. What is the hill for artificial gas for cooking and 
lighting if a family used 8350 cu. ft. and if the cost is 
$ 1.15 per thousand cubic feet? 

5. What is the cost of : 

a. 48 qt. of milk @ ^ per quart ? 

b. 54 lb. of nuts @ 331 ^ per pound ? 

c. 72 lb. of meat @ 16| ^ per pound ? 









IV. DENOMINATE NUMBERS 


71. Review the tables of measures, if necessary, given at 
the back of the book. Units of Measure, by which corre¬ 
sponding quantities are measured, are given in these tables. 
The Measure of a concrete quantity is a number of units 
of measure of the same kind. 

Thus, a line may be eight feet in length, — the measure being 
eight and the unit one foot. Again, a lot may contain 2345 square 
feet, — the measure being 2345 and the unit one square foot. 

72. Expressions like 8 ft., 2345 sq. ft., 3 gal. +1 qt., etc. 
are called Denominate Numbers. 

73. Expressing Denominate Numbers in Terms of Smaller 
Units. 

Example 1. Express 3 A. 50 sq. rd. in terms of square 
rods. 

Solution . 1. 3 A. = 3 x 160 sq. rd. = 480 sq. rd. 

2. .*. 3 A. 50 sq. rd. = 480 sq. rd. + 50 sq. rd. = 530 sq. rd. 

Example 2. Express T. in pounds. 

Solution. & T. = & x 2000 lb. = 320 lb. 

Example 3. Express | rd. in terms of smaller units. 

Solution. 1. | rd. = | x 5^ yd. 


2 . 

3. 

4. 


= i X ¥ yd. = -V- yd. = 4| yd. 
f yd. = f X 3 ft. = 2f ft. 


| ft. = § x 12 in. = 8 in. 


f rd. = 4 yd. 2 ft. 8 in. 
108 


DENOMINATE NUMBERS 


109 


EXERCISE 54 

Reduce: 

1. 5 mi. to rods ; to feet. 

2. 12 gal. to quarts ; to pints. 

3. 15 sq. yd. to square feet; to square inches. 

4. 1 wk. to hours ; to minutes. 

5. 3 T. to hundredweight; to pounds. 

6. 8 bu. to quarts ; to pints. 

7. 15° to minutes ; to seconds. 

8 . 3 cu. yd. to cubic feet; to cubic inches. 

9. 4 sq. rd. to square yards ; to square feet. 

10. 24 rd. to yards ; to feet. 

11 . i cu. yd. to cubic feet; to cubic inches. 

12. J mi. to rods ; to feet. 

13. 35 rd. to yards ; to feet. 

14. to minutes ; to seconds. 

15. bu. to pecks ; to quarts ; to pints. 

16. ji- A. to square rods ; to square feet. 

17. f T. to hundredweight; to pounds. 

18. 9.5 bu. to pecks ; to pints. 

19. .56 wk. to hours ; to minutes. 

20. 8.6 sq. rd. to square feet. 

21. 3.6 gal. to quarts ; to pints. 

22. 12 gal. 3 qt. to pints. 

23. 1 T. 15 cwt. 81 lb. to pounds. 

24. 3 A. 5 sq. rd. to square yards. 

25. 365 da. 5 hr. to hours. 

26. 2 mi. 253 rd. to feet. 

27. 8 bu. 6 pk. to quarts. 


110 


NEW HIGH SCHOOL ARITHMETIC 


28. 3 yd. 2 ft. 5 in. to inches. 

29. 4 gal. 31 qt. to pints. 

30. 5 sq. yd. 7 sq. ft. to square feet. 

31. 15 cu. yd. 8 cu. ft. to cubic feet. 

Express the following in terms of smaller units as in 
Example 3, page 108. 

32. | gal. 35 -|| rd. 38. | rd. 41. ^ yr. 

33. bu. 36. 2.8125 lb. 39. .625 bu. 42. A mi. 

34. 2.51 hr. 37. .75 wk. 40. T. 43. \ rd. 


74. Expressing Denominate Numbers in Terms of Larger 
Units. 

Example 1. Express 5870 ft. in rods. 

Solution. 1. Since 16^ ft. = 1 rd., 

2; 5870 ft. = 5870 4 - 16± 

3. = 5870 x -fa 

4. = 355f| rd. 

5. = 355.7- rd. 

The approximate decimal form of the result is probably 
the most useful form. 


Example 2. Express 1523 ft. in terms of larger units 
with integral multipliers. 


Solution. 1. Since 3 ft. = 1 yd., 
1523 ft. = 1523 4- 3 = 507 yd. 2 ft. 

2. Since 5£ yd. = 1 rd. 

507 yd. = 507 4 - 5£ 

= 92 rd. 1 yd. 

3. .-. 1523 ft. = 92 rd. 1 yd. 2 ft. 


.1523 = 507 1 yd. 

= 507 yd. 2 ft. 
507 51 

= 507 x 2 
= = 92 t 2 x rd. 

= 92 rd. 1 yd. 


Note. — This process has long been called Reduction Ascending. 




DENOMINATE NUMBERS 


111 


EXERCISE 55 

Express in common and also in decimal fractional form : 

1. 3 pk. as part of a bushel. 

2. 7 pt. as part of a gallon. 

3 . 35 min. as part of an hour. 

4 . 250 rd. as part of a mile. 

5 . 90 sq. in. as part of a square foot. 

6. 15 cu. ft. as part of a cubic yard. 

7 . 100 sq. rd. as part of an acre. 

8. 1000 yd. as part of a mile. 

9 . 30,000 sq. ft. as part of an acre. 

10. 20 in. as part of a yard. 

11 . 3500 ft. as part of a mile. 

12. 1350 lb. as part of a ton. 

13 . 75 rd. as part of a mile. 

14 . 250 da. as part of a year. 

15 . 30 wk. as part of a year. 

16 . 14 in. as part of a yard. 

17 . 175 rd. as part of a mile. 

18 . 120 sq. rd. as part of an acre. 

19 . 30 in. as part of a yard. 

20. 1400 lb. as part of a ton. 

Express in fractional form and also in decimal form cor¬ 
rect to two decimal places as in Example 1. 

21. 1259 rd. as miles. 

22. 7843 lb. as tons. 

23 . 15,786 ft. as miles. 

24 . 278 qt. as gallons. 

25 . 73,840 sq. ft. as acres. 

26 . 10,000 ft. as miles. 


112 


NEW HIGH SCHOOL ARITHMETIC 


27 . 3500 yd. as miles. 

28 . 23,450.1b. of wheat as bushels of wheat. 

(See Tables of Measures for the number of pounds in a bushel.) 

29 . 31,275 lb. of oats as bushels of oats. 

30 . 16,870 lb. barley as bushels of barley. 

31 . 480 qt. as bushels. 

32 . 9365 cu. ft. as cubic yards. 

33 . 15,000 ft. as miles. 

34 . 7500 yards as miles. 

35 . 10,000 doz. as gross. 

36 . 6500 da. as years. 

37 . 30,000 lb. as tons. 

38 . 150,000 sq. ft. as acres. 

39 . 15 cwt. 6 lb. as tons. 

40 . 2 pk. 2 qt. as part of a bushel. 

41 . 16 hr. 25 min. as hours. 

42 . 2 qt. 1 pt. as part of a gallon. 

43 . 8 ft. 11 in. as feet. 

44 . 4 yd. 1 ft. 2| in. as part of a yard. 

45 . 71 lb. 12 oz. as part of a hundredweight. 

46 . 3 pk. 1 qt. as part of a bushel. 

47 . 3 qt. 1 pt. as part of a gallon. 

48 . 4 da. 20 hr. as part of a week. 


Express in terms of larger units as in Example 2. 


49 . 

975 pt. dry measure. 

56 . 

3758 lb. 

50 . 

2357 oz. avoirdupois. 

57 . 

5432 yd. 

51 . 

2609 pt. liquid measure. 

58 . 

2654 ft. 

52 . 

45,432 in. 

59 . 

5000 yd. 

53 . 

113,292 sq. ft. 

60 . 

8500 ft. 

54 . 

80,000 seconds of time. 

61 . 

27,000 lb. 

55 . 

98,582 sq. ft. 

62 . 

15,000 yd. 


DENOMINATE NUMBERS 


113 


75. Adding and Subtracting Denominate Numbers. 

Example 1. Add 17 lb. 10 oz., 12 lb. 9 oz., and 8 lb. 
7 oz. 

Solution . 1. 17 lb. + 10 oz. 

12 lb. + 9 oz. 

8 lb. + 7 oz. 

37 lb. + 26 oz. 

2. But 26 oz. = 1 lb. + 10 oz. 

3. 37 lb. + £6 oz. = 38 lb. + 10 oz. 

Example 2. Subtract 3 mi. 201 rd. from 6 mi. 76 rd. 

Solution. 1. 6 mi. + 76 rd. = 5 mi. + 396 rd. 

3 mi. + 201 rd . = 3 mi. + 201 rd. 

2 mi. + 195 rd. 

In this solution, 1 mi. of the 6 mi. is changed to 320 rd. and 
added to the 76 rd., making 396 rd. 

EXERCISE 56 

Find the sum of: 

1. 12 gal. 3 qt. 1 pt., 7 gal. 1 qt., and 24 gal. 2 qt. 1 pt. 

2. 4 cwt. 67 lb., 11 cwt. 49 lb., and 2 cwt. 83 lb. 

3. 34 bu. 3 pk. 7 qt., 46 bu. 1 pk. 5 qt. 1 pt., 55 bu. 

2 pk. 6 qt., and 27 bu. 3 pk. 1 pt. 

4 . 71 da. 22 hr. 18 min., 36 da. 16 hr. 48 min., 60 da. 

11 hr. 32 min., and 49 da. 9 hr. 23 min. 

5. 45° 38' 40", 123° 17' 33", 78° 44' 55", and 65° 46' 12". 

6. 7 mi. 163 rd., 2 mi. 313 rd. 5 yd., 9 mi. 36 rd. 4 yd. 
6 mi. 3 yd., and 244 rd. 2 yd. 

7. 13 T. 11 cwt., 8 T. 12 cwt. 64 lb., 5 T. 8 cwt. 74 lb., 

12 T. 36 lb., and 14 cwt. 63 lb. 

8. 165 cu. ft. 828 cu. in., 23 cu. ft. 562 cu. in., and 37 
cu. ft. 983 cu. in. 

9. 3 sq. mi. 288 A., 543 A. 19 sq. rd., 74 A. 53 sq. rd., 
and 610 A. 149 sq. rd. 





114 


NEW HIGH SCHOOL ARITHMETIC 


10. 12 mi. 98 rd. 5 yd., 8 mi. 212 rd. 4 yd., 156 rd. 5 yd., 
18 mi. 313 rd. 4 yd., 36 mi. 286 rd., and 3 mi. 125 rd. 5 yd. 

Subtract: 

11. 2 bu. 3 pk. 7 qt. from 13 bu. 2 pk. 5 qt. 

12. 49° 52' 41" from 87° 39' 25". 

13 . 16° 37' from 97° 42' 15". 

14 . 38 gal. 3 qt. 1 pt. from 48 gal. 3 qt. 

15. 42 lb. 7 oz. from 52 lb. 3 oz. 

16 . 6 cu. yd. 26 cu. ft. 752 cu. in. from 18 cu. yd. 9 cu. ft, 

17 . 7 mi. 16 rd. from 15 mi. 38 rd. 

18 . 14 T. 16 cwt. from 23 T. 11 cwt. 

19 . 9 A. 147 sq. rd. from 15 A. 

20. 1 da. 20 br. from 5 da. 15 hr. 48 min. 

21. 4 A. 152 sq. rd. from 78 A. 153 sq. rd. 

22. The sides of a field are 8 rd. 1 yd. 2 ft., 9 rd. 1 yd. 

2 ft., 12 rd. 1 yd., and 10 rd. 1 yd. 1 ft. What is the length 
of a fence around it ? 

23 . To trim a girl’s garment there are needed 2 pieces of 
lace 13 inches long, one piece li yd. long, and one piece 3| 
yd. long. How much lace is required for the garment ? 

24 . A man agreed to build 115 rd. of fence. On the first 
day, he built 27 rd. 3 yd.; and the second, 32 rd. 5 yd.; 
and on the third day, 26 rd. 1 yd. How much remained 
to be built ? 

25 . From a barrel containing 50 gal. of lubricating oil, a 
dealer sold the following quantities: 5 gal. 3 qt., 3 gal., 
1 qt., 5 pt., 5 gal., 2\ gal., 7 qt., 2 gal. 1 pt., 3 gal. 2 qt. f 

3 pt. 

How much should be left in the barrel ? 


DENOMINATE NUMBERS 


115 


76. Finding the Time between Two Dates. 

Example 1. Find the exact number of days from the 13th 
of June to the 10th of October of the same year. 


Solution. 1. There remain in June 17 da. 

There are in July 31 da. 

There are in August 31 da. 

There are in September 30 da. 
There are in October 10 da. 

2. the total number of days = 119 da. 


Note. — The first of the given dates is not included, but the 
last is included. 

Example 2. Express in years, months, and days the time 
between Apr. 12, 1929, and Nov. 20, 1932. 

Solution. 1. From Apr. 12, 1929, to Apr. 12, 1932, is 3 year. 

2. From Apr. 12, 1932, to Nov. 12,1932, is 7 mo. 

3. From Nov. 12, 1932, to Nov. 20, 1932, is 8 da. 

4. .*. the total time is 3 yr. 7 mo. 8 da. 

This method of solution is useful when the exact time in years, 
months, and days is wanted. 

Note. — When reckoning from the last day of a month having 
31 days, the number of months from then until the last day of the 
month preceding the final date is taken as the number of months, 
and then the number of days is counted. 

Thus, the time from Mar. 31 to Oct. 17 is found as follows: 

From Mar. 31 to Sept. 30 is 6 months. 

From Sept. 30 to Oct. 17 is 17 days. 

Hence, the whole time is 6 months and 17 days. 

Example 3. How long is it from Feb. 13, 1929, to May 6, 
1932? 

Yr. Mo. Da. Yr. Mo. Da. 

Solution. May 6, 1932 = 1932 5 6 = 1932 4 36 

Feb. 13, 1929 = 1929 2 13 = 1929 2 13 

3 2 23 

Since 13 da. cannot be subtracted from 6 da., 1 mo. of the 5 mo. 
in the minuend is changed to 30 days and added to the 6 da., mak¬ 
ing 36 da. 



116 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 57 

Find the exact number of days from : 

1. Mar. 23 to Dec. 11 of the same year. 

2. Apr. 12 of one year to Jan. 18 of the next year. 

3 . Sept. 4 to Dec. 29 of the same year. 

4 . July 7 to May 25 of the next year. 

5 . Nov. 22 to Aug. 9 of the next year. 

6. Jan. 17 to Sept. 26 of the same year. 

7 . Feb. 3 to Oct. 16 of the same year. 

8. Sept. 22 to Mar. 12 of the next year. 

9. May 3 to Nov. 24 of the same year. 

10. Apr. 26 to Feb. 19 of the next year. 

Find, as in Example 2, p. 115, the time in years, months, 
and days from: 

11. March 13, 1933, to Jan. 28, 1938. 

12. Feb. 17, 1933, to June 8, 1938. 

13 . April 26, 1933, to March 24, 1935. 

14 . Aug. 21, 1933, to May 15, 1937. 

15 . Dec. 13, 1933, to Dec. 4, 1935. 

16 . Oct. 28, 1933, to Mar. 16/1939. 

17 . Jan. 12, 1933, to Apr. 4, 1939. 

18 . Feb. 17, 1933, to June 6, 1943. 

19 . Nov. 6, 1933, to Feb. 15, 1937. 

20. Oct. 23, 1933, to May 1, 1937. 

21 - 30 . Find, as in Example 3, p. 115, the time between 
the dates given in Examples 11 to 20, and compare your 
results with those found in Examples 11 to 20. 


DENOMINATE NUMBERS 


117 


77. Multiplying Denominate Numbers. 

Example. Multiply 2 mi. 135 rd. 5 yd. by 4< 

Solution. 1. 2 mi. 135 rd. 5 yd. 

_4_ 

8 mi. 540 rd. 20 yd. 

2. But 20 yd. = 3 rd. + 3* yd. 

.*. the distance = 8 mi. 543 rd. 3^ yd. 

3. 543 rd. = 1 mi. 223 rd. 

.*. the distance = 9 mi. 223 rd. 3| yd. 

EXERCISE 58 

Multiply: 

1 . 13 bu. 3 pk. by 3. 8 . 15 gal. 3 qt. 1 pt. by 6. 

2 . 8 ft. 7 in. by 5. 9 . 18 mi. 275 rd. 5 yd. by 4. 

3 . 4 gal. 3 qt. by 3. 10 . 5 da. 19 hr. 37 min. by 10. 

4 . 7 sq. ft. 32 sq. in. by 8. 11 . 16 T. 8 cwt. 85 lb. by 5. 

5 . 5 rd. 4 yd. by 4. 12 . 37 cu. ft. 560 cu. in. by 11. 

6 . 7 bu. 3 pk. 5 qt. by 2. 13 . 12 lb. 7 oz. by 25. 

7 . 26° 38' by 3. 14 . 16 rd. 1 yd. 2 ft. by 9. 

15 . 5 A. 87 sq. rd. by 10. 

16 . How many feet of wood are required for 26 pieces 
each 3' 4" in length ? 

17 . How many feet of iron rod are needed to make 185 
marking stakes, each 1' 8 " in length ? 

18 . Find the value of 35 A. 68 sq. rd. of land at $ 225 
per acre. 

19 . What is the value of 18,240 square feet of land at 
$ 750 per acre ? 

20. How much concrete is required to make 150 concrete 
stepping stones each containing 875 cu. in. of concrete ? 

21. How many yards of carpet must be bought to give 5 
strips, each 4 yd. 2 ft. 8 in. in length ? 



118 


NEW HIGH SCHOOL ARITHMETIC 


78. Dividing Denominate Numbers by an Integer or a 
Fraction. 

Example. Divide 47 A. + 51 sq. rd. by 3. 

Solution. 1. 47 A. -4- 3 = 15 A. + remainder of 2 A., or 320 
sq. rd. 

2. 320 sq. rd. + 51 sq. rd. =371 sq. rd. 

371 sq. rd. h- 3 = 123f sq. rd. 

3. .*. (47 A. + 51 sq. rd.)-n 3 = 15 A. + 123f sq. rd. 


EXERCISE 59 


Divide: 


1 . 

2 . 

3. 

4. 

5 . 
11 . 
12 . 

13 . 

14 . 

15 . 

16 . 

17 . 

18 . 

19 . 

20 . 
21 . 


35 bu. by 3. 

53 gal. 3 qt. by 5. 

11 ft. 10 in. by 4. 

22° 30' by 4. 

135 A. 80 sq. rd. by 2. 

234 T. 17 cwt. 10 lb. by 10. 
35 bn. 3 pk. 5 qt. by 3. 

10' 9" by 3. 

16 rd. 91 yd. by 15. 

3 wk. 5 da. 4 br. by 4. 
Multiply 13 lb. 10 oz. by 3J. 


6 . 

41 

lb. 4 oz. by 5. 

7 . 

17 

T. 5 cwt. by 3. 

8 . 

25 

sq. yd. 3 sq. ft. by 6. 

9 . 

19 

rd. 21 yd. by 2. 

10 . 

22 

A. 152 sq. rd. by 6. 


-11'9- 


Divide 2 T. 2 cwt. by f. 

Multiply 7 mi. 210 rd. 5 yd. by 4. 

Multiply 12 cu. ft. 25 cu. in. by \. 

Multiply 38° 58' 16" by f. 

A carpenter is asked to build 
four cold-frame sashes to cover the 
cold frame pictured. What will be 
the outside dimensions of each sash ? 

22. The distance between the first floor and the second 
floor of a house is 10' 6". If the stairway is to contain 17 
steps (including the landing on the second floor), how much 
rise is there in each step ? 


i 









DENOMINATE NUMBERS 119 


79. Dividing a Denominate Number by a Denominate 
Number. 


Example 1. 

Solution. 1. 

2 . 

3. 


Divide 12 yd. 2 ft. by 1 yd. 1 ft. 9 in. 

12 yd. 2 ft. = 38 ft. = 456 in. 

1 yd. 1 ft. 9 in. = 57 in. 

456 in. -f- 57 in. = 8. 


Example 2. What part of 7 gal. 2 qt. is 3 gal. 3 qt.? 

Solution. 1. By § 53, the part is used as the dividend and the 
whole as the divisor. 

2- 7 gal. 2 qt. = 30 qt. 

and 3 gal. 3 qt. = 15 qt. 

.*. the fractional part = I'LSb — 1. 

30 qt. 2 

Hence 3 gal. 3 qt. is | of 7 gal. 2 qt. 


EXERCISE 60 

Divide: 

1. 15 bu. by 3 pk. 

2. 54 gal. by 4 gal. 2 qt. 

3 . 18 T. by 2 T. 5 cwt. 

4 . 171 gal. 3 qt. 1 pt. by 3 gal. 1 pt. 

5 . 6 T. 10 cwt. 84 lb. by 1 T. 12 cwt. 71 lb. 

6. 502 da. 21 hr. 45 min. by 4 da. 3 hr. 45 min. 

7. 10 mi. 165 rd. 2\ yd. by 224 rd. 2 yd. 

What part of 

8. 5 da. is 1 da. 13 hr. 30 min.? 

9. A steel rod 15 ft. long is put in an automatic machine 
to be worked up into pieces 2\ in. in length. If 1 in. is 
allowed for cutting on each piece, how many pieces will be 
obtained from the rod? 

10. How many bricks each containing 64 cu. in. of mate¬ 
rial can be made from 50 cu. ft. of the raw material ? 



120 


NEW HIGH SCHOOL ARITHMETIC 


11. About bow many concrete blocks each containing 900 
cu. in. of concrete can be made from 1 cu. yd. of concrete? 

12. How many pieces of tape 30 ,f long can be cut from a 
10-yd. bolt of tape ? 

13 . If an estate of 460 acres is divided equally among 6 
children, how many acres and square rods will each child 
receive ? 

14 . The capacity of a tank is 31 cu. ft. 528 cu. in. How 
many barrels will it contain if the capacity of a barrel is 
2 cu. ft. 1152 cu. in. ? 

15 . What is the rate of a train in feet per minute which 
is traveling 38 miles per hour ? 

16 . What is the value of a carload of wheat weighing 
8 T. 17 cwt. at % .63 per bushel if a bushel of wheat 
weighs 60 lb.? 

17 . A man burned 9 T. 13 cwt. of coal in his furnace 
between Nov. 8 and Feb. 17 inclusive. How much did he 
burn per day ? 

18 . Sound travels about 1085 ft. per second. How long 
will it take the sound of a gun to reach a point 21 miles 
away? 

19 . It takes a train 9 sec. to go 528 ft. (a) How long 

will it take it to go one mile ? ( p ) How many miles will 

it travel in one hour ? 

20. If the moon revolves around the earth in 27 da. 7 hr. 
42 min., over what part of its orbit does it travel in 1 da.? 
Express the result decimally to two places. 

21. If a steel rail weighs 63 lb. per yard, how many tons 
of rail will be required to lay a mile of single-track railroad ? 

22. If a cubic foot of water weighs 62-| lb. and lead is 
11.44 times as heavy as water, how many cubic inches of 
lead are there in a piece weighing 43 lb. 8 oz.? 


DENOMINATE NUMBERS 


121 


23 . If a cubic foot of water weighs 1000 oz. and iron is 
7.6 times as heavy as water, what is the weight of 405 cu. 
in. of iron ? 

24 . If a railroad company is asked to put in a switch 
track one quarter mile in length, how many 60-ft. rails are 
needed ? 

25 . Find the cost at $ 23.75 per ton of a load of hay 
weighing 4435 lb., if the weight of the empty wagon is 
1670 lb. 

26 . A farmer delivered at the mill loads weighing net: 
2875 lb., 2962 lb., 2885 lb., 3018 lb., 2992 lb., 3084 lb., 
3115 lb., and 2925 lb. 

Find the amount due him if oats were selling at 27 $ per 
bushel. (See Tables for number of pounds in a bushel.) 

27 . A millman shipped a car containing 54,912 lb. of 
wheat. What was due him if wheat was selling at $ .65 
per bushel ? 

28 . He also shipped 45,228 lb. of rye, which was sold at 
$ .47 per bushel. What was due him ? 

29 . Find the value, at $ .58 per bushel, of the following 
five loads of wheat weighing: 

3175 lb.; 3064 lb.; 3220 lb.; 2985 lb.; 3043 lb. 

30 . A boy bought at a mill the following order of grains 
for chicken feed: 

125 lb. of bran at $ 1.85 per hundredweight. 

125 lb. of middlings at $ 2.10 per hundredweight. 

75 lb. of corn meal at $ 1.65 per hundredweight. 

25 lb. of oil meal at $ 2.25 per hundredweight. 

100 lb. shelled corn at $ .38 per bushel. 

100 lb. oats at $ .26 per bushel. 

Find the total of his bill. 


122 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 61 
Miscellaneous Examples 

1. In 1931 French aviators broke the record for duration 
of a flight by remaining in the air about 75 hr. and 20 min., 
and covering a distance of 5470 mi. What was the average 
distance flown in one hour ? 

2. How far from the earth, in miles, is an aviator who 
is up 11,000 ft. ? 

3 . How far will a projectile go in a minute, if it is 
moving 2040 ft. per second? 

4 . What is the total cost at $ 12.35 per ton of the fol¬ 
lowing loads of coal: 

4720 lb.; 4564 lb.; 4488 lb.; 4675 lb.; 4810 lb. ? 

5 . Change 2.32 A. to square rods. 

6. a. Change 18,000 ft. to miles. 

b. Express 18,000 ft. in terms of higher units. 

7 . A barrel of gasoline, containing 541 gallons, is con¬ 
sumed in the gas machine of a country home in about 7 
weeks, a. What is the amount consumed per day ? 
b. What is the expense per day if the gasoline costs $ .43-1- 
per gallon ? 

8. A tobacco grower delivered at the warehouse three 
loads of tobacco weighing respectively 3278 lb., 3795 lb., 
and 4015 lb. What was the value if tobacco was selling 
for $ 3.25 per hundredweight ? 

9 . Measure the diameter of a silver dollar. If 1,000,000 
silver dollars were placed in a line, how far would they 
reach ? 

10. It was estimated that 11,590,000,000 gal. of milk 
were produced in the United States in 1915. How much 
was this per capita if the population was approximately 
100,000,000? 


DENOMINATE NUMBERS 


123 


SUPPLEMENTARY TOPICS 

Whether or not the following topics should he taught, 
depends upon local conditions. 

A. Metric Measures. B. Foreign Money. 

C. Apothecaries’ Weights. 


A. The Metric System of Measures 

80. The Metric System of Measures is employed exclu¬ 
sively now in many foreign countries. It is also used in 
the sciences, and in certain of the Departments of the 
United States Government. 

Many of our manufacturers must make articles for 
foreigners according to directions given in metric units. 

The system was first adopted in France in 1799. 


Measures of Length 


81. The fundamental unit of length is the Meter. This 
was intended to be one ten-millionth of a quadrant of the 
meridian of the earth running through Paris. Compared 
with the English units of length, one meter equals about 
39-37 inches. 

Table 


10 millimeters (mm.) 
10 centimeters 
10 decimeters 
10 meters 
10 dekameters 
10 hektometers 
10 kilometers 


= 1 centimeter (cm.) 
= 1 decimeter (dm.) 
= 1 meter (m.) 

= 1 dekameter (Dm.) 
= 1 hektometer (Hm.) 
— 1 kilometer (Km.) 
= 1 miriameter (Mm.) 


Equivalents 


1 meter = 39.37 in. 

1 yard = .914 m. 


1 kilometer = about f mile. 
1 mile = 1.6093 Km. 


124 


NEW HIGH SCHOOL ARITHMETIC 


From the table it is clear that there are in a meter 10 
decimeters, 100 centimeters, and 1000 millimeters. 



The prefixes of the word “ meter ” always have the same 
meaning: 

thus, milli means one thousandth of the unit; 

centi means one hundredth of the unit; 
deci means one tenth of the unit; 

Deka means ten times the unit; 

Hecta means one hundred times the unit; 

Kilo means one thousand times the unit; 

Miria means ten thousand times the unit. 


Measures of Surface 


Table 


100 square millimeters (sq. mm.) — 1 square centimeter (sq. cm.) 


= 1 square decimeter (sq. dm.) 

= 1 square meter (sq. m.) 

= 1 centar (Ca.) 

= 1 square dekameter (sq. Dm.) 
= 1 ar (a.) 

= 1 square hektometer (sq. Hm.) 

— 1 hektar (Ha.) 

— 1 square kilometer (sq. Km.) 
The square meter is used in measuring ordinary surfaces; the 

square kilometer, in measuring the areas of countries; the ar and 
hektar, in measuring land. 


100 square centimeters 
100 square decimeters 

100 square meters, or centars 

100 square dekameters, or ars 

100 square hektometers 


Equivalents 


1 sq. cm. = .155 sq. in. 

1 sq. m. = 1.196 sq. yd. 
1 sq. Km. = .3861 sq. mi. 
la. = 3.954 sq. rd. 

1 Ha. = 2.471 A. 


1 sq. in. = 6.452 sq. cm. 
1 sq. yd. = .8361 sq. m. 
1 sq. rd. = .2529 a. 

1 sq. mi. = 2.59 sq. Km. 
1 A. = .4047 Ha. 













DENOMINATE NUMBERS 


125 


Measures of Contents 

Table 

1000 cubic millimeters ( cu. mm.) = 1 cubic centimeter (cu. cm.) 
1000 cubic centimeters = 1 cubic decimeter (cu. dm.) 

1000 cubic decimeters = 1 cubic meter (cu. m.) 


Equivalents 

1 cu. m. = 1.308 cu. yd. 1 cu. yd. = .7645 cu. m. 


Measures of Capacity 

Table 


10 milliliters (ml.) 
10 centiliters 
10 deciliters 
10 liters 
10 dekaliters 
10 hektoliters 


= 1 centiliter (cl.) 
= 1 deciliter (dl.) 
= 1 liter (l.) 

— 1 dekaliter (Dl.) 
= 1 hektoliter (HI.) 
= 1 kiloliter (Kl.) 


Equivalents 


1 liter = 1.0567 liq. qt. 
1 liter = .9081 dry qt.. 
1 hektoliter = 3.531 cu. ft. 

1 hektoliter = 26.417 gal. 


1 liquid qt. =.9463 1. 

1 dry qt. = 1.101 1. 

1 gallon = 3.785 1. 

1 bushel = .3524 HI. 


Note. — The liter is used in measuring liquids and small fruit; 
the hektoliter, in measuring grain, vegetables, and liquids in 
casks. 


Measures of Weight 

The principal unit of weight is the Gram. It is the 
weight of a cubic centimeter of distilled water at 39.2° 
Fahrenheit. 


126 


NEW HIGH SCHOOL ARITHMETIC 


Table 


10 milligrams = 
10 centigrams = 
10 decigrams = 
10 grams = 
10 dekagrams = 
10 hektograms = 
1000 kilograms = 


1 centigram (eg.') 

1 decigram ( dg.) 
1 gram (g.) 

1 dekagram (Dg.) 
1 hektogram ( Hg.) 
1 kilogram (Kg.) 
1 metric ton (T.) 


Equivalents 

1 kilogram = 2.2046 lb. avoirdupois 1 lb. avoirdupois = .4536 K g. 

Note. —The gram is used in weighing letters, precious metals, 
and jewels, and in mixing medical prescriptions; the kilogram, in 
weighing ordinary articles; the metric ton, in weighing heavy 
articles. 


EXERCISE 62 

1. By holding the edge of a ruler against the scale on 
p. 124, determine the approximate number of millimeters 
in 3 in.; also in 1.25 in. 

2. Similarly determine the approximate English length 
equivalent to 45 mm.; also to 93 mm. 

3. Measure the diameter of a penny in millimeters. 

4. Measure the diameter of a five-cent piece in milli¬ 
meters. 

5. Measure the length and the width of your school 
room in the metric system. 

(If your class does not possess a meter stick, the teacher 
in charge of Physics classes in your school certainly will 
have one that you can use.) 

6. Class Exercise. Represent upon the blackboard a 
square meterr. Inside it, draw a square representing 
a square yard. 


DENOMINATE NUMBERS 


127 


82. Reducing Denominate Numbers in the Metric System. 

In the tables of measures of length, weight, and capac¬ 
ity, any unit is ten times the next smaller unit of the table. 
Hence to change any number of these units to the next 
smaller units, merely move the decimal point one place to 
the right; to change to the next larger unit, move the 
decimal point one place to the left. 

Thus, 25 m. = 250 dm. = 2500 cm. = 25000 mm.; 
and 25 m. = 2.5 Dm. = .25 Hm. = .025 Km. 

Expressions like 2 m. 5 dm. 3 mm. are therefore unneces¬ 
sary in the metric system, for 

2 m. 5 dm. 3 mm. = 2000 mm. 500 mm. 3 mm. 

= 2503 mm. 

In changing units of surface or square measure, the 
decimal point must be moved two places to the right or 
left for every change to a smaller or a larger unit respect¬ 
ively, for in the table of surface measure, each unit is one 
hundred times the next smaller unit. 

Similarly, in changing cubic units, the decimal point must 
be moved three places. 


EXERCISE 63 

Express: 

1. 4.72 m. as centimeters ; as millimeters ; as kilometers, 

2. .073 Kl. as liters ; as centiliters. 

3. 8.325 Kg. as grams ; as decigrams ; as milligrams. 

4. 73.85 Km. as meters ; as dekameters. 

5. .0548 sq. Hm. as square meters. 

6. 3 m. 5 dm. 4 cm. as centimeters; as meters. 

7. 7 Kg. 6 Dg. 5 g. as grams. 

8. 5 1. 4 dl. 6 ml. as liters. 

9. 78.56 m. as centimeters ; as hectometers. 

10 . 735 cu. m. as cubic decimeters. 


128 


NEW HIGH SCHOOL ARITHMETIC 


11. Add 78.26 mg., 8.53 dg., 3 g. (Change all of them to 
milligrams.) 

12. Add 47.6 m., 83.05 Dm., .45 Km. 

13 . Add 314 sq. m., 5.62 sq. m., and .043 sq. Dm. 

14 . Subtract 89.7 Dg. from 2.816 Kg. 

15 . Subtract 2.482 mm. from .5016 m. 

16 . Subtract 60.4 dl. from 4.307 HI. 

17 . Multiply 5.03 m. by 2.5 and express the result in 
hectometers. 

18 . Multiply .8259 sq. dm. by 35 and express the result 
in square meters. 

19 . Multiply 47.7 mg. by .6 and express the result in 
centigrams. 

20. Divide 93.98 ml. by 25.4 dl. 

21. Divide .0735 Kg. by 2.3 and express the result to 
three decimal places in grams. 

22. Divide 4.686 Hm. by 825 m. 

23 . Divide .0602784 cu. cm. by .644 and express the 
result in cubic millimeters. 

24 . Multiply 56.82 dm. by 4.3. 

25 . Divide 4.7835 Km. by 635 m. 

83. Only very simple problems involving conversion of 
metric units to English units or vice versa are likely to 
occur. 

Example 1. Express 15.3 Km. in miles. 

Solution. 1. Since 1 Km. = .6214 mi. 

2. 15.3 Km. = 15.3 x .6214 mi. 

= 9.5+ mi. 

Example 2. Express 1 ton in kilograms. 

Solution. 1. Since 1 lb. = .4536 kg. 

2. .-. 2000 lb. = 2000 x .4536 kg. = 907.2 kg. 


DENOMINATE NUMBERS 


129 


EXERCISE 64 

Express: 

1. 1 yd. in decimeters. 

2. 1 pk. in dekaliters. 

3 . 1 sq. ft. in square decimeters. 

4 . 1 cu. m. in cubic feet. 

5 . 1 centar in square feet. 

6. 1 pt. (liquid) in deciliters. 

7 . 31 Hm. in rods. 

8. 288 sq. ft. in square meters. 

9 . 5 yd. 2 ft. in meters. 

10. 20 cu. ft. in cubic meters. 

11. 25 lb. 14 oz. in kilograms. 

12. 12 bu. 3 pk. in dekaliters. 

13 . 3 mi. 200 rd. in kilometers. 

14 . 3500 kilograms in pounds. 

15 . 2750 kilometers in miles. 

16 . 15 hectars in acres. 

17 . 40 acres in hectars. 

18 . 60 lb. (1 bu. wheat) in kilograms. 

19 . 31i gal. (1 bbl.) in kiloliters. 

20. 15 Km. in miles. 

21. A merchant bought 300 liters of wine at 83 1 per liter 
and sold it at $ 5.00 per gallon. What were his profits ? 

22. A merchant bought silk at $ 1.95 per meter and sold 
it at $3.00 per yard. How much did he gain on 100 
meters ? 

23 . A merchant wants about 500 yards of a certain grade 
of silk. How many meters should he order ? \ 

24 . A wine merchant received 1000 liters of wine. How 
many quart bottles will he be able to fill ? 


130 


NEW HIGH SCHOOL ARITHMETIC 


25 . Foreign letters from the United States weighing 
15 g. or less may he mailed for 5 ^ postage. What is the 
equivalent weight in English units ? 

26 . How many kiloliters of gasoline are there in a barrel 
containing 54.5 gallons ? 

27 . A French grain dealer ordered through a broker 
10,000 kilograms of wheat. How many bushels should the 
broker purchase ? 

28 . A broker must fill an order for 25,000 kilograms of 
oats. How many bushels must he order ? 

29 . What is the equivalent in French units of 100,000 
gallons of gasoline ? 

30. What is the equivalent in French units of 50,000 
pounds of beef ? 

31 . On Sept. 2, the English troops continued “ their 
attacks on a front of 45 Km.” (From official war report.) 
What is the English equivalent of this distance ? 

B. Foreign Money 

84. English Money 

4 farthings (far.) = 1 penny (d.') 

12 pence = 1 shilling ( 5 .) 

20 shillings = 1 pound (£) 

Is. = $.243+; 1 £ = $4.8665 

French Money 

100 centimes = 1 franc (fr.) = $ .193. 

German Money 
100 Pfennigs = 1 Mark (M)= $.238. 

Mexican Money 
100 centavos = 1 peso = $.498. 


DENOMINATE NUMBERS 


131 


EXERCISE 65 

1. Change 1 £ to pence. 

2. Change 100 d. to pounds. 

3. Express 75 s. in higher denominations. 

4. Change 10,000 centimes to francs. 

5. How much United States money should a soldier re¬ 
ceive for 75 fr. ? 

6. How many francs should a traveler receive for $ 150, 
if no charge is made for the exchange ? 

7. How much United States money will a traveler re¬ 
ceive in exchange for 75 £ ? 

8. A United States soldier bought in France articles cost¬ 
ing 3 fr. 50 cen.; 2 fr. 40 cen.; 5 fr. 85 cen. and 75 cen. 
How much did he spend ? 

9. Express f £ in shillings ; in pence. 

10. Express .45 £ in shillings ; in pence. 

11. Express .78 fr. in centimes. 

12. Express -|fr. in centimes. 

13. How much change should be received from a 5-pound 
note offered in payment of purchases amounting to S£ 17 s. 
10 d.? 

14. An American traveler going to England exchanges 
$ 1000 in United States money for English money. How 
much should he receive ? 

15. What will be the cost of 15 doz. towels at 18 s. 6 d. 
per dozen ? 

16. a. The European Edition of the New York Herald in 
the summer of 1918 sold for 20 centimes per copy. How 
much is the equivalent in American money? 

b. The annual subscription rate outside of Paris was 
75 fr. per year. How much is this in American money ? 


132 


NEW HIGH SCHOOL ARITHMETIC 


C. Apothecaries , Measures 

85. Apothecaries’ measures, liquid and dry, are given in 
the collection of tables at the back of the book. 

Reduction, both ascending and descending, and other 
operations with these measures are performed as with other 
measures. 


EXERCISE 66 

1 . How many scruples are there in 1 pound ? 

2 . Change 8 § 5 3 to drams. 

3. Change 2 lb 3 § 7 3 to drams. 

4. Change 75 gr. to drams. 

5. Change 250 gr. to ounces. 

6 . Change 96 3 to ounces; to drams. 

7. Change 1 pt. to fluid drams. 

8 . Change 1± gal. to fluid ounces. 

9. Change 26/3 to fluid ounces. 

10 . Change 4240 minims to pints. 

11 . Express in higher units 8027 gr. 

12 . Express in higher units 35,798 minims. 

13. Express in higher units 951,000 3 . 

14. Add 8§ 5 3 , 10§ 73 3s, 6323 , and 6|43 9 3 . 

15. Erom 3 fb 11 §, take 15 3 . 

16. Add 30 1 5f% and IO 8 /§. 

17. Erom 1 gal. take 5 O 12/§. 

18. How many bottles of hand lotion each containing 
3/5 can be filled from 3 quarts of the lotion ? 

19. Erom 140 4take 90 13 J%. 

20. How much quinine is required to fill 15 dozen 3-grain 
capsules ? 


DENOMINATE NUMBERS 


133 


21 . How much of each of the following drugs must be 
taken to make enough for 24 powders, each of which is to 
contain the following amounts: 

Extract nux vomica, 6 gr.; caffeine citrate, 24 gr.; sodium 
salicylate, 13 ; sodium phosphate, 13 ? 


V. INVOLUTION AND EVOLUTION 


86 . If a number is used as a factor two or more times, 
the result is called a Power of the number. 

Thus, 3 x 3 or 9 is the second power of 3 ; 

and 5 x 5 x 5 or 125 is the third power of 5. 

5x5x5 is written 5 3 . The small number written at 
the right and a little above the 5 is called an Exponent; 
the number 5 is called the Base. 

An integral Exponent indicates the number of times that 
the base is used as a factor. 

Thus, 2 5 means 2 x 2 x 2 x 2 x 2 or 32. 

87. Involution is the process of raising a number to a 
required integral power. 

The second power of a number is called the Square of the 
number. 

Thus, 7 squared or 7 2 is 49. 

The third power of a number is called the Cube of the 
number. 

Thus, 4 cubed or 4 8 is 4 x 4 x 4 or 64. 


EXERCISE 67 

Find the value of : 


1 . 

52. 

6 . 10 3 . 

11. 10 4 . 

16. 

II 2 . 

2. 

3 3 . 

7. 3 4 . 

12. 7 3 . 

17. 

4 4 . 

3. 

6 2 . 

8. 5 4 . 

13. 92. 

18. 

35. 

4. 

2 4 . 

9. 6 3 . 

14. 252. 

19. 

26. 

5. 

82. 

10. 20 2 . 

15. 30 3 . 

20. 

6 4 . 


134 


INVOLUTION AND EVOLUTION 


135 


* 1 - a) 2 - 

22. (|)3. 

23. a) 2 . 

24. 

25. 

26. 


(I) 3 * 

(f) 2 - 

© 2 . 


27. (1)4. 

28. ( t 4 x ) 2 . 

29. 12 3 . 

30. 85 2 . 

31. (2.5) 2 . 

32. ( 1 . 1 ) 2 . 


33. ( 3 . 5 ) 2 . 

34. ( 4 . 3 ) 2 . 

35. ( 1 . 5 ) 2 . 

36. ( 1 . 3 ) 2 . 

37. (.25)2. 

38. (. 04 ) 2 . 


39. (.21)2. 

40. (2.03)2. 

41. (. 003 ) 2 . 

42. (3.12)2. 

43. (20.13)2. 

44. (.215)2. 


45. Memorize the square of each of the integers 1, 2, 3, 
... etc., up to 20 inclusive. 

46. Determine and memorize the cube of each of the 
integers 1, 2, 3, ... etc., up to 10 inclusive. 

88 . If one number is the square of a second, then the 
second is called the Square Root of the first. 

Thus, since 25 = 5 2 , then 5 is the square root of 25. 

If one number is the cube of a second, then the second is 
called the Cube Root of the first. 

Thus, since 8 = 2 3 , then 2 is the cube root of 8. 

89. Evolution is the process of finding a required root of 
a number. 


90. The Radical Sign, V , when written over a nunlber, 
indicates some root of the number. The particular root 
wanted is indicated by writing an Index to the left of and a 
little above the radical sign. 

\/8 means the cube root of 8, or 2. 

^81 means the fourth root of 81. This is 3. 

When no index is written, then the square root is 
indicated. 

Thus, V25 means the square root of 25, or 5. 

91. Finding Roots by Inspection. — Certain square and cube 
roots' can be given at sight if pupils have done as directed 
in Examples 45 and 46 of Exercise 67. 


136 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 68 


Give: 

1. +81. 

2 . -m. 

3. Vli4. 

4. +27 


5. +121. 

6 . + 1000 , 

7. +196. 

8. +125. 


9. +256. 

10. +512. 

11. +400. 

12. +343. 


13. +324 

14. +729. 

15. +361. 

16. +2l6. 


17. Find +£§. 

Solution. = ~, for ^ x ^ 

’25 V25 5 55 


16 

25 


18. Find -v'^. 


Solution. 



^8 

\025 



2 2 
^x- x 
o 5 


2 

5 


8 

125* 


Find similarly: 

19. Vf. 25. +p|. 31. -V^.37. 43. 

20. V||. 26. Vi||. 32. +++ 38. ++*. 44. +pf. 

21 - '^T5T- 27 - +iW 33 ‘ " v 4'V 39 ‘ VjV- 45 ' VM- 

22. +p. 28. +^. 34. +*. 40. +^&. 46. VSJ. 

23. +ff 29. +ff. 35. 41. VW- 47. </&. 

24. +Ip. 30. +J. 36. V^. 42. +ip. 48. +pf. 


92. Computing the actual or approximate Square Root is 
based upon the following method of squaring a number. 

To find 74 2 . 

Notice that 74 = 70 + 4. To multiply 70 + 4 by 70 +4, one 
may first multiply by 70 and then by 4 as below. 

70 x (70 + 4) = 70 2 + 70 x 4 
4 x (70 + 4) = 70 x 4 + 4 2 


Hence (70 + 4) 2 = 70 2 + 2 x 70 x 4 + 4 2 . 








INVOLUTION AND EVOLUTION 


137 


That is, to square a number of two figures, first square the 
tens, add twice the product of the tens and the units and 
finally add the square of the units. 

Thus, 23 2 = 20 2 + 2x 20 x3 + 3 2 = 400 + 120 + 9 = 529. 

93. Notice next that l 2 = 1, 10 2 = 100, 100 2 = 10,000 ; etc. 

From these results it is clear that 

a. The square of a number of one figure has either one or 
two figures in it; 

b. The square of a number of two figures has either three or 
four figures in it; 

c. The square of a number of three figures has either five or 
six figures in it. 

Therefore, working backward, if a power be divided into 
groups of two figures each, beginning with the units 7 figure, 
for each such group there will be one figure in the square 
root. The groups are called Periods. 

Thus, 6889 becomes 68 89. The square root must have in it two 
figures. It cannot have three figures by (c) above, and it must 
have more than one by ( a ) above. 

Similarly 45,632 becomes 4 56 32. Its square root must have in 
it three figures, a hundreds’, a tens’, and a units’ figure. 

If there are figures to the right of the decimal point, di¬ 
vide them also into periods of two figures each, starting from 
the decimal point. 

Thus, 4357.265 becomes 43 57.26 50. The last period on the 
decimal side is completed by annexing a zero to the given number. 
The square root of this number must have two figures to the left 
of the decimal point and two to the right of it. 

94. The Process of Computing the Square Root. 

Example 1. Find the square root of 6889. 

Solution. 1. 6889 divided into periods becomes 68 89. There 
are in the square root a tens’ and a units’ figure. 

2. The tens’ figure must be 8, for 80 2 = 6400 and 90 2 = 8100. 


L38 


NEW HIGH SCHOOL ARITHMETIC 


That is, the square root must be 80 plus a number x less than 10. 

6889 must = 80 2 + 2 x 80 X x + x t% 

= 80 3 + x(2 x 80 + x) 68 89 

From 6889, subtract 80 2 or 6400. 64 00 

Then 2 x 80 x x + x 2 must be 4 89 

To find x, form the trial divisor 2 x 80 or 160. This is con¬ 
tained in 489 three times with a remainder. This suggests that 
x = 3. 

To the trial divisor 160 add 3, making 163. This is the com* 
plete divisor. 3 x 163 exactly equals 489. 

The square root then is 80 + 3 or 83. This can be proved by 
squaring 83. 

This solution is arranged in the following form. 

80 + 3 


68 89 
64 00 


Trial divisor 2 x 80; 160 

489 - 4 - 160 = 3+ Add 3. __3 

Complete divisor 163 


Hence the square 
root is 83. 


4 89 


Note. - 
solution. 


- The zeros crossed out are usually omitted in the 


Rule. — To find the square root of a number: 

1. Separate the number into periods. (§ 93) 

2 . Find the greatest square number in the leftmost period; 
write its square root in the root; subtract the square itself from 
the leftmost period, and to the result annex the next period. 

3 . Form the trial divisor by doubling the root already found 
and annexing zero to the result. 

4 . Divide the remainder in Step 2, by the trial divisor. Annex 
the quotient to the root already found, and add the quotient to 
the trial divisor, forming the complete divisor. 

5 . Multiply the complete divisor by the root figure last ob¬ 
tained and subtract the product from the remainder of Step 2. 

6 . If other periods remain, repeat Steps 3, 4, and 5 until there 
is no longer a remainder, or until the desired number of decimal 
places has been obtained for the root. 








INVOLUTION AND EVOLUTION 


139 


Note. 1. — It sometimes happens that, on multiplying a com- 
plete divisor, the product is greater than the remainder. In such 
cases, the figure of the root last obtained is too great, and the 
next smaller integer must be substituted for it. 

Note 2. — If the remainder, in Step 4, is less than the trial 
divisor, the next figure of the root is zero. Annex zero to the 
root, and bring down the next period. Then start again with 
Step 3. 


Example 2. Find the square root of 4944.9024. 

70.32 


Solution. The first trial divisor is 140. 
Since this is more than 44, the first remainder, 
annex zero to the root, obtaining 70. Then 
bring down the next period, 90. 

The second trial divisor is then 1400. Divid¬ 
ing 4490 by it, gives 3 as the next figure of the 
root. 


49 44.90 24 


49 



1400 

3 

44 90 

1403 

42 09 

14060 

2 

2 81 24 

14062 

2 81 24 


Example 3. Find the square root of 12, correct to three 


decimal places. 


Solution. The solution adjoining is accord¬ 
ing to the rule. 

The square root to three places is 3.464. 


3.464 


12 00 00 00 
9 


60 

_4 

64 


3 00 


2 56 


680 

6 

686 

6920 

4 

6924 


44 00 


41 16 


2 84 00 


2 76 96 


7 04 


EXERCISE 69 


Find the square root of: 

X. 1849. 3. 3136. 5. 8281. 7. 55,225. 

2. 3844. 4. 5476. 6. 10,609. 8. 264,196 



















140 


NEW HIGH SCHOOL ARITHMETIC 


9. 104,329. 

10. 14.3641. 

11. 2735.29. 

12. 33.1776. 


13. .521284. 

14. 6811.2009. 

15. 44,605.44. 

16. 24.7009. 


17. 3344.3089. 

18. 5838.4881. 

19. 224,486.44. 

20. .256036. 


Eind the approximate square root of the following, cor¬ 
rect to three decimal places : 


21. 

7. 

25. 

17.3. 

29. 

72.05. 

33. 125.48. 

22. 

23. 

26. 

2.85. 

30. 

643.27. 

34. 3410.73. 

23. 

35. 

27. 

37.2. 

31. 

58.908. 

35. 6714.95. 

24. 

47. 

28. 

145.63. 

32. 

375.625. 

36. 18,956.241. 

95. 

Tables 

are 

employed 

to 

avoid the 

computation of 


Powers and of Roots of numbers. On p. 141 is a table of the 
squares, cubes, square roots, and cube roots of all numbers 
from 1 to 100 inclusive. In the column headed by the 
letter N is the number. Opposite the number, in the ap¬ 
propriate column is found its square, its cube, its square 
root, or its cube root. 

Thus, in the second column of numbers is found the number 
64. In the first column to the right is the square of 64, namely 
4096. In the second column to the right, is the cube of 64, 
namely 262,144. In the next column to the right, is the square 
root, namely 8. And in the last column, is the cube root of 64. 
namely 4. 

EXERCISE 70 

Determine from the table, or with the help of the table : 


1 . 

39 2 . 

9. 

V56. 

17. 

-Vn. 

25. 

■m. 

33. 

m 

2. 

76 3 . 

10. 

V70. 

18. 

V37. 

26. 

V-8-fi- 

V 64* 

34. 


3. 

58 2 . 

11. 

86 2 . 

19. 

■\/28. 

27. 

Vff 

35. 

Vff. 

4. 

47 2 . 

12. 

■\/99. 

20. 

SI 

28. 


36. 

Vjf 

5. 

29 3 . 

13. 

23 3 . 

21. 

my- 

29. 

vk- 

37. 

■m 

6. 

V38. 

14. 

-^59. 

22. 

my- 

30. 

•v/6 3 

V 6T* 

38. 


7. 

^82. 

15. 

V 65. 

23. 

my- 

31. 

v«. 

39. 


8. 

V9l. 

16. 

46 s . 

24. 

my- 

32. 


40. 

m 


INVOLUTION AND EVOLUTION 141 


N. 

Squares 

Cubes 

Square 

Roots 

Cube 

Roots 

1ST. 

Squares 

Cubes 

Square 

Roots 

Cube 

Roots 

1 

1 

1 

1 . 

1 

51 

2,601 

132,651 

7.141 

3.708 

2 

4 

8 

1.414 

1.259 

52 

2,704 

140,608 

7.211 

3.732 

3 

9 

27 

1.732 

1.442 

53 

2,809 

148,877 

7.280 

3.756 

4 

16 

64 

2. 

1.587 

54 

2,916 

157,464 

7.348 

3.779 

5 

25 

125 

2.236 

1.709 

55 

3,025 

166,375 

7.416 

3.802 

6 

36 

216 

2.449 

1.817 

56 

3,136 

175,616 

7.483 

3.825 

7 

49 

343 

2.645 

1.912 

57 

3,249 

185,193 

7.549 

3.848 

8 

64 

512 

2.828 

2 

58 

3,364 

195,112 

7.615 

3.870 

9 

81 

729 

3. 

2.080 

59 

3,481 

205,379 

7.681 

3.892 

10 

100 

1,000 

3.162 

2.154 

60 

3,600 

216,000 

7.745 

3.914 

11 

121 

1,331 

3.316 

2.223 

61 

3,721 

226,981 

7.810 

3.936 

12 

144 

1,728 

3.464 

2.289 

62 

3,844 

238,328 

7.874 

3.957 

13 

169 

2,197 

3.605 

2.351 

63 

3,969 

250,047 

7.937 

3.979 

14 

196 

2,744 

3.741 

2.410 

64 

4,096 

262,144 

8. 

4. 

15 

225 

3,375 

3.872 

2.466 

65 

4,225 

274,625 

8.062 

4.020 

16 

256 

4,096 

4. 

2.519 

66 

4,356 

287,496 

8.124 

4.041 

17 

289 

4,913 

4.123 

2.571 

67 

4,489 

300,763 

8.185 

4.061 

18 

324 

5,832 

4.242 

2.620 

68 

4,624 

314,432 

8.246 

4.081 

19 

361 

6,859 

4.358 

2.668 

69 

4,761 

328,509 

8.306 

4.101 

20 

400 

8,000 

4.472 

2.714 

70 

4,900 

343,000 

8.366 

4.121 

21 

441 

9,261 

4.582 

2.758 

71 

5,041 

357,911 

8.426 

4.140 

22 

484 

10,648 

4.690 

2.802 

72 

5,184 

373,248 

8.485 

4.160 

23 

529 

12,167 

4.795 

2.843 

73 

5,329 

389,017 

8.544 

4.179 

24 

576 

13,824 

4.898 

2.884 

74 

5,476 

405,224 

8.602 

4.198 

25 

625 

15,625 

5. 

2.924 

75 

5,625 

421,875 

8.660 

4.217 

26 

676 

17,576 

5.099 

2.962 

76 

5,776 

438,976 

8.717 

4.235 

27 

729 

19,683 

5.196 

3 J 

77 

5,929 

456,533 

8.774 

4.254 

28 

784 

21,952 

5.291 

3.036 

78 

6,084 

474,552 

8.831 

4.272 

29 

841 

24,389 

5.385 

3.072 

79 

6,241 

493,039 

8.888 

4.290 

30 

900 

27,000 

5.477 

3.107 

80 

6,400 

512,000 

8.944 

4.308 

31 

961 

29,791 

5.567 

3.141 

81 

6,561 

531,441 

9. 

4.326 

32 

1,024 

32,768 

5.656 

3.174 

82 

6,724 

551,368 

9.055 

4.344 

33 

1,089 

35,937 

5.744 

3.207 

83 

6,889 

571,787 

9.110 

4.362 

34 

1,156 

39,304 

5.830 

3.239 

84 

7,056 

592,704 

9.165 

4.379 

35 

1,225 

42,875 

5.916 

3.271 

85 

7,225 

614,125 

9.219 

4.396 

36 

1,296 

46,656 

6. 

3.301 

86 

7,396 

636,056 

9.273 

4.414 

37 

1,369 

50,653 

6.082 

3.332 

87 

7,569 

658,503 

9.327 

4.431 

38 

1,444 

54,872 

6.164 

3.361 

88 

7,744 

681,472 

9.380 

4.447 

39 

1,521 

59,319 

6.244 

3.391 

89 

7,921 

704,969 

9.433 

4.464 

40 

1,600 

64,000 

6.324 

3.419 

90 

8,100 

729,000 

9.486 

4.481 

41 

1,681 

68,921 

6.403 

3.448 

91 

8,281 

753,571 

9.539 

4.497 

42 

1,764 

74,088 

6.480 

3.476 

92 

8,464 

778,688 

9.591 

4.514 

43 

1,849 

79,507 

6.557 

3.503 

93 

8,649 

804,357 

9.643 

4.530 

44 

1,936 

85,184 

6.633 

3.530 

94 

8,836 

830,584 

9.695 

4.546 

45 

2,025 

91,125 

6.708 

3.556 

95 

9,025 

857,375 

9.746 

4.562 

46 

2,116 

97,336 

6.782 

3.583 

96 

9,216 

884,736 

9.797 

4.578 

47 

2,209 

103,823 

6.855 

3.608 

97 

9,409 

912,673 

9.848 

4.594 

48 

2,304 

110,592 

6.928 

3.634 

98 

9,604 

941,192 

9.899 

4.610 

49 

2,401 

117,649 

7. 

3.659 

99 

9,801 

970,299 

9.949 

4.626 

50 

2,500 

125,000 

7.071 

3.684 

100 

10,000 

1,000,000 

10. 

4.641 
































142 


NEW HIGH SCHOOL ARITHMETIC 


SUPPLEMENTARY TOPIC 

96. Using Powers and Roots in Formulas. 

A rule of computation can be expressed in words or by 
means of a Formula in which letters represent numbers. 

Thus, the cost of several articles of the same kind at the same 
price for each is found by multiplying the number of articles by 
the price of each. Briefly: 

Cost = number of articles x price of each. 

If we let C = the cost, n = the number of articles, and 
p = the price, then 

C=n x p 

is the shortest way of expressing this rule. 

If n = 30 and p — 5 then C = 30 x 5/, or 150^, or 
$ 1.50. 

Again, wages = number of hours x pay per hour. 

If we let W = the wages, n = the number of hours, and 
p = the pay per hour, this rule is written briefly thus : 

W=n x p. 

If, now, n = 6 and p — 40 W = 6 x 40^, or 240 or 

$ 2.40. 

Many formulas have in them powers of roots. 

Example 1. S = 16 x t 2 ft. is the rule for finding how 
many feet an object will fall in t seconds. 

Suppose t — 15. How much is S ? 

Solution. 1. 5=16 xl5 2 . 

2. From the table, 15 2 = 225 ; .-. S = 16 x 225. 

3. v. S = 3600 ft. 

Example 2. t = 4 sec. is the rule for finding the num¬ 
ber of seconds it will take an object to fall S feet. 

Suppose S — 1225. How much is t ? 

Solution. 1. t = \ V1225. 

2. On page 141, opposite 35 2 is 1225 in the column of squares. 

3. Since 35 2 = 1225, then VI225 is 35. 

4* .-. t = 1 x 35 = 8f seconds. 


INVOLUTION AND EVOLUTION 


143 


EXERCISE 71 

1. Review Example 1, page 142, to make certain that you 
know what the formula S = 16 t 2 means. 

By this formula, find how far an object will fall in: 

a. 12 seconds 6. 18 seconds c. 21 seconds 

2 . Review Example 2, page 142, to make certain that 
you know what the formula t = means. 

By this formula, find how long it will take an object to 
fall: 

a. 841 ft. b. 729 ft. c. 2304 ft. 

3. The formula A = 3.14 x r 2 is the rule for finding the 
area of a circle which has the radius r. 

By this formula, find the area of the circle of which the 
radius is: 

a. 13 ft. b. 24 in. c. 32 in. d. 44 in. 

4. The formula F=| X 3.14 x r 3 is the rule for finding 
the volume of the sphere which has the radius r. 

By this formula, find the volume of the sphere of which 
the radius is : 

a. 8 in. b. 13 in. c. 16 in. d. 21 in. 

I _ 

5. The formula r = - V it A is the rule for finding the 

7r 

radius of the circle of which the area is A. Let w = 

By this formula, find the radius of the circle of which: 
a. A = 154 sq. in. b. A — 616 sq. in. c. A — 1386 sq. in. 

6 . By the formula a = V H 2 — 6 2 , find a if H = 35 and 
6 = 21. 

Suggestion. By the table, 35 2 = ?; 21 2 = ? .*. H 2 — b 2 = ? 

Now find a, by using the table. 

7. By the formula of Example 6, find a if: 

a. #=45, 6 = 27 6. #=39,6 = 15 

8. By the formula F=% mv 2 , find F if : 

a. m = 1500, v = 65 6. m = 1800, v = 73 



144 


NEW HIGH SCHOOL ARITHMETIC 


9. By the formula s = -y/ V, find s if: 
a. F= 95 b. V= 59,319 c. V= 314,432 


10. By the formula r = find the 


value of r if 


V = 38,808 and tt = ^ 


11. By the formula R = yj—, find R if A = 21,952,000 
and P = 8000. 

12. By the formula A — P(RY, find A if P = $ 1500 and 
R = 1.1. 


VI. PRACTICAL GEOMETRY AND APPLICATIONS 


97. Geometrical Figures are formed of points, lines, and 
surfaces. Four kinds of lines are studied in elementary 
geometry: ^ 

a. Straight Lines like line l. - -- 

One and only one straight line can be drawn through two 

points. 

A straight line is supposed to extend indefinitely in each 
of two directions. 

The part of a straight line between two A B 

points of it is called a Segment; as seg- * • 

ment AB. 

The shortest distance between two points is measured 
along the segment between them. Segments are measured 
by units of length, as the yard, the foot, the mile, etc. 

The part of a straight line on one side of 
a point of it is called a Ray, as ray m. 

b. Broken Lines like EFGH. They 
consist of segments of straight lines. 

c. Curved Lines like CD. They do not 
have any parts that are straight. 

d. Closed Lines, like those adjoining. 

Simple closed lines in a plane 

surface divide the surface into 
two parts, one inside the line 
and one outside. • 


P 

F 


G 


F' 




98. An Angle is the figure formed by 
two rays drawn from the same point; 
as Angle AOB. o 



345 







146 


NEW HIGH SCHOOL ARITHMETIC 


The common point is the Vertex of the angle, and the 
two rays are the Sides of the angle. 

A common angle is the Right Angle. 

Whenever an object has a “square 
corner,” two of its edges form a right 
angle. 

Two lines that form a right angle 
are Perpendicular ; as lines TS and BS. 

Angles are measured by a unit called a 
Degree, —an angular degree. An angular 
degree is one ninetieth of a right angle. 

An Acute Angle is an angle less than 
a right angle ; as angle ABC. 

An Obtuse Angle is an angle greater than 
a right angle and less than two right angles ; 
as angle DEF. 

99. Two lines are Parallel if they lie in the same plane 

and do not meet, no matter how far A _ B 

they are extended. The distance be¬ 
tween parallel lines is always the same. G D 



EXERCISE 72 

1. Mention some objects that have a surface or surfaces 
that are practically plane surfaces. 

2. Mention one or more objects inclosed by a spherical 
surface. 

3. Mention an object that has some Other curved surface. 

4. Point out in the room some parallel lines. 

5. Point out in the room some perpendicular lines. 

6. What kind of angle "is formed by the hands of a 
clock at 3 o’clock ? How many degrees are there in it ? 

7. Answer questions like those in Example 6 when the 
hands point to: a. 2 o’clock; b. 4 o’clock; c. 5 o’clock. 








PRACTICAL GEOMETRY AND APPLICATIONS 147 


8. How do carpenters mark a straight line between two 
points by means of a string ? 

9. Draw a straight line. Through a point on it, draw 
a perpendicular to it, using the square corner of a card as a 
pattern right angle. 

10. Draw a straight line. Place a point on your paper 
but not on the line. Through this point draw a perpen¬ 
dicular to the line, using the corner of the card as a pattern 
right angle. 

11. Draw a straight line at least two inches long. On 
this line, place two points, at least one inch apart. At 
each of these points draw a perpendicular to the line. 
What kind of lines do the two perpendiculars appear to be ? 

12. Draw two lines that are parallel. At a point of one 
of them draw a perpendicular. What kind of angles does 
this perpendicular form with the other parallel ? 

13. Can you fold a piece of paper so as to form a straight 
line ? 

14. Can you now fold the paper so that the first crease 
will be perpendicular to the second ? 

15. Following the suggestion of Example 11, can you, by 
folding a piece of paper three times, make two parallel 
creases in the paper ? 

100. A Polygon is a closed broken line lying in a plane; 
as ABCDE. AB, BC, etc., are called its Sides; points A, 
B , C, etc., are called its Vertices. A B 

line, as AC, joining two non-consecutive 
vertices is a Diagonal of the polygon. / \ 

The Perimeter of the polygon is the A<A 
sum of the lengths of its sides. Nv s' 

Note. — Formerly it was customary to B 

say that a polygon is a part of a plane bounded by a broken line? 
and then to call that line the perimeter of the polygon. This 


148 


NEW HIGH SCHOOL ARITHMETIC 


statement is still to be found in elementary texts although it 
does not agree with the best modern mathematical practice. 

Since a polygon is a closed line in a plane, it incloses a 
part of the plane called the interior of the polygon. 

This interior of the polygon is measured by units of sur¬ 
face measure. The Area of the polygon is the number of 
times that the interior of the polygon contains the unit of 
surface measure. 

Thus, if a polygon is such that its interior contains 8 sq. ft., 
then the area is 8 and the unit of measure is a square foot. 



101. Kinds of Polygons. 

a. A Triangle is a polygon having three sides; as tri¬ 
angle ABC. 

Any side of a triangle may be con¬ 
sidered its Base; as BC. The perpen¬ 
dicular to the base from the opposite 
vertex is the Altitude to that base; as 
AD. 

h. A Quadrilateral is a polygon having four sides. 

c. A Parallelogram is a quadrilateral whose opposite sides 

are parallel ; as ABCD. b e G 

It is proved in geometry that the i -] 7 

opposite sides of a parallelogram are j j / 
equal. _ j / 

Any side of a parallelogram may be A F D 

considered its Base, and the perpendicular between this side 
and the one opposite is then the Altitude; as base AD and 
altitude EF. b c 

d. A rectangle is a special kind of par¬ 
allelogram. Its angles are all right angles. 

Any side, as AD , may be considered its A D 

Base, and then either of the adjoining sides is its Altitude. 

e. A Square is another special kind of parallelogram. It 







PRACTICAL GEOMETRY AND APPLICATIONS 149 


is also a special kind of rectangle. Its angles are all right 
angles, and its sides are all equal. 

EXERCISE 73 

1. Draw as nearly as you can a rectangle whose base is 
4 inches and whose altitude is 2 inches. 

2. Draw as nearly as you can a large parallelogram. 
Measure its opposite sides. Are they approximately equal ? 

3. Draw a triangle, each of whose angles is an acute 
angle. 

a. Select one side as base, and to that side draw the 
altitude. (See Example 10, Exercise 72.) 

b. Similarly draw the altitudes to the second and third 
sides. 

c. What do you observe about the altitudes ? 

4. Draw a triangle which has for one of its angles a 
right angle. What sides of the triangle may conveniently 
be selected for base and altitude ? 

5. Draw a triangle which has for one of its angles an 
obtuse angle. Select one of the sides of the obtuse angle 
as the base, and to that side draw the altitude. 

6. Draw a parallelogram. Select one side as base, and 
to that side draw the altitude of the parallelogram. 

7. Represent a rectangular lot whose length is 120 ft. and 
width 30 ft. In your drawing, let 30 ft. be represented by 
1 in. (This is called drawing to the scale 1 in. = 30 ft.) 

8. Represent a square field whose sides are 100 ft. long, 
letting 1 in. equal 25 ft. 

9. Draw to the scale ^ in. = 1 ft. a rectangle represent¬ 
ing a kitchen 14 ft. 6 in. wide and 12 ft. 3 in. long. 

10. Draw to the scale -§- in. = 1 ft. a rectangle represent¬ 
ing the outside foundation walls of a house 33 ft. wide and 
38 ft. long, if the walls are 15 in. in thickness. 


150 


NEW HIGH SCHOOL ARITHMETIC 


102. Area of a Rectangle. — In the figure at the right, let 
the sides of square U each represent 1 inch. Then AB 

represents 2 inches and BC 5 inches. A _ D 

U represents one square inch. ABCD 

contains 2 x 5 or 10 square inches. That 
is, the area of ABCD is 10. 

How many and what kind of square units would there 
he in ABCD : 

a. if AB were 3 inches and BC were 5 inches ? 

b. If AB were 5 feet and BC were 8 feet ? 

c. If AB were 7 yards and BC were 12 yards ? 

The base and altitude are measured by the same unit of 
length; then the interior of the rectangle is measured by 
a square whose side is this same unit of length. It is clear 
then that the number of these square units within the rec¬ 
tangle is the product of the number of units of length in 
the base by the number in the altitude of the rectangle. 
This fact is abbreviated commonly in the 

Rule. — The area of a rectangle is the product of its base and 
altitude. 

103. If a represents the number of linear units in the 
altitude, and b the number in the base, and A the number 
of square units in the interior of the rectangle, the rule can 
be expressed in the form 

A = a x b. 

A rule expressed in this manner is called a Formula. 
This is the formula for the area of a rectangle. 

Instead of writing the cross, X, for the sign of multipli¬ 
cation between numbers represented by letters, the letters 
are written side by side. Thus a x b is written ab. There¬ 
fore the formula for the area of a rectangle is A = ab. 

Example. Find the area of a rectangle whose base is 
15 ft. and whose altitude is 9 ft. 


a 



PRACTICAL GEOMETRY AND APPLICATIONS 151 


Solution. 1. The formula is A = ab. 

2. Substitute 15 for a and 9 for b. Then 

A = IS X 9 = 135 sq. ft. 

EXERCISE 74 

1. What is the area of a square whose side is : 

a. 9 ft.? b. 12 yd.? c. 20 rd.? d. 3.75 yd.? 

2. What is the area of a rectangle whose : 

a. altitude is 16 ft. and whose base is 33 ft. ? 

b. altitude is 6| ft. and whose base is 121 ft. ? 

c. altitude is 4*' 10" and whose base is 12' 4" ? 

3. What is the altitude of a rectangle whose area is 
544 sq. ft. and whose base is 10 rd. ? 

4. Determine the length of the side of a square whose 
area will be: 

a. 196 sq. yd. b. 400 sq. ft. c. 625 sq. rd. 

5. Determine the approximate length of the side of a 
square whose area will be : 

a. 150 sq. ft. b. 200 sq. yd. c. 1000 sq. ft. 

6. What is the base of a rectangle whose area is 3^ sq. 
yd. and whose altitude is 3 ft. 6 in. ? 

7. What is the number of acres in a square field whose 
side is 396 ft. ? 

8. What part of an acre is a lot 100 ft. by 300 ft. ? 

9. What part of an acre is a lot 60 ft. by 120 ft. ? 

10. What will be the cost of a flexo-tile floor for a hotel 
kitchen 15 ft. by 35 ft. at $ .45 per square foot? 

11. What will be the cost for the canvas for the floor of 
a sleeping porch 10 ft. by 18 ft. at $ .75 per square yard ? 

12. What is the value at $ 185 per acre of a field 24 rd. 
wide and 35 rd. long ? 


152 


NEW HIGH SCHOOL ARITHMETIC 


13. A firm builds a warehouse four stories high. The 
interior dimensions of the building are 50 ft. by 200 ft. 
How many square feet of floor space have they provided 
for themselves ? 

14. What will be the expense for a cement sidewalk on 
the two sides of a corner lot 50 ft. by 160 ft., if the walk is 
5 ft. wide, at 12 $ per square foot? 

15. What will be the cost of covering a bathroom floor 
7 ft. 6 in. by 9 ft. 8 in. with flexo-tile at $ .45 per square foot ? 

16. What will it cost to cover the floor of a sun parlor 
10 ft. 10 in. by 18 ft. 4 in. with tile at $ .60 per square foot ? 

17. Approximately how many bricks 8 inches long and 
4 inches wide, will be required to lay a sidewalk 5 ft. wide 
and 275 ft. long? 

18. Determine the cost at $ 13.50 per thousand of the 
bricks for paving an alley 545 ft. long and 12 ft. wide. 
The bricks are laid so that the exposed surface is 8 in. by 

in. Assume that the contractor can buy only whole 
thousands of bricks. 

19. How many thousand paving stones 9 in. by 4 in. will 
be needed for 2 miles of road 20 ft. wide ? 

20. A contractor offers to lay an asphalt pavement at 
$3.60 per square yard. The street is 50 ft. wide. How 
much must each lot owner be assessed for each foot in the 
width of his lot ? 

21. Determine the cost of painting a house 27 ft. wide 
and 33 ft. long, having an average height of 22 ft. at $ .30 
per square yard. (No allowance to be made for windows.) 

22. Determine the cost of painting a billboard 20 ft. 
wide and 10 ft. high at $ .25 per square yard. 

23. One gallon of paint is enough for two coats of paint 
for about 250 sq. ft. of surface. How many gallons of 
paint must a farmer order for a barn 30 ft. x 50 ft., averag¬ 
ing 23 ft. in height ? 


PRACTICAL GEOMETRY AND APPLICATIONS 153 


104. Problems of the Plastering Contractor.—Plastering, 
both inside and outside, is figured by the square yard of 
surfa-ce covered. The price charged includes furnishing 
and placing in position the lath upon which the plaster is 
spread, as well as the plaster work proper. Usually the 
contractors do not make any allowance for openings for 
doors or windows, as it costs as much to work around these 
openings as it would to cover them. 

EXERCISE 75 

1. What will it cost to plaster the walls and ceiling of 
a room 14 ft. by 22 ft., 9 ft. high, at $ .40 per square yard ? 

2. Eind the cost of plastering the ceiling and walls of a 
town hall 30 ft. wide, 45 ft. long, and 16 ft. high at $ .45 
per square yard. 

3. What will be the cost of the stucco finish for a garage 
18 ft. wide, 22 ft. long, and averaging 10 ft. in height at 
$ .55 per square yard ? 

4. What will be the cost of re-plastering the walls of a 
kitchen 12 ft. wide, 14 ft. long, and 9 feet high at $ .30 per 
square yard ? 

5. What will be the contract price for the plastering of 
the walls and ceilings of a house, having the following 
rooms, at $ .45 per square yard ? 

Kitchen, 14 ft. by 14 ft., 9 ft. high. 

Living room, 20 ft. by 14 ft. 6 in., 9 feet high. 

Dining room, 13 ft. by 13 ft. 6 in., 9 ft. high. 

Hall, 12 ft. by 14 ft. 6 in., 9 ft. high. 

Pantry, 5 ft. by 12 ft., 9 ft. high. 

Bedroom A, 12 ft. by 12 ft., 8 ft. 6 in. high. 

Bedroom B, 13 ft. by 17 ft., 8 ft. 6 in. high. 

Bedroom C, 12 ft. 6 in. by 7 ft. 6 in., 8 ft. 6 in. high. 

Bathroom A, 7 ft. by 8 ft., 8 ft. 6 in. high. 

Bathroom B, 6 ft. by 11 ft., 8 ft. 6 in. high. 


154 


NEW HIGH SCHOOL ARITHMETIC 


105. Papering. — Wall paper is approximately 18 in. wide 
and comes in single rolls of 8 yd. or double rolls of 16 yd. 
The single rolls therefore contain 4 sq. yd. of paper. 
Paper comes also in other widths, but in each case there 
are approximately 4 sq. yd. in a single roll of domestic 
paper. When estimating the quantity of paper needed, 
allowance must be made for the waste in cutting to match 
the pattern. It is customary to cut full length strips 
where there are not openings, — in the expectation that the 
small spaces over doors and windows can be covered with 
the pieces of the rolls remaining after the long strips are 
cut. 

A rough estimate of the quantity needed is obtained by 
assuming that a single roll will cover 33 sq. ft. of surface. 
The total surface of the room, less the openings, divided by 
33 will give the number of rolls. 

Another way is to subtract the total width of all doors 
and windows from the perimeter of the room, and divide 
the result by the width of the paper. This will give the 
number of strips needed for the walls. The number that 
can be cut from a roll depends upon the height of the 
ceiling. 

Example. What will be the cost at $ .60 per roll of the 
paper for the walls of a room 17 ft. by 12 ft. if the ceiling 
is 9 ft. high, and for the ceiling at $ .45 per roll. The 
paper is 18 in. wide and is bought in double rolls. There 
is an 8 in. baseboard running around the room. There is 
one group of windows 9 ft. wide, another 7 ft. wide, one 
small door 3 ft. 6 in. wide and one double doorway 7 ft. 
6 in. wide. 

Solution. 1. Each strip must be 9' —8" long, or 8' 4" in length. 

2. .-. 5 strips can be cut from a double roll. (48' -4- 8' 4" = 5+). 

3. The total width of all openings is 7' -f- 7' 6" + 9' + 3' 6" 
or 27'. 

4. The perimeter of the room is 17' + 17' + 12' + 12' or 58'. 


PRACTICAL GEOMETRY AND APPLICATIONS 155 


5. Subtracting the width of all openings leaves 31'. 

6. The number of strips needed = 31'-f-18" = 20f or 21 strips. 

7. Since only 5 strips can be cut from a double roll, it will 
take 4 double rolls and one single roll, — all together 9 rolls. 

8. The cost will be 9 x $ .60 or $5.40 for the sides. 

9. On the ceiling, if the strips are run the narrow way of the 
room, the number of strips needed = 17' -f-18"= 11 * or 12 strips. 

10. Each strip must be 12' long. 4 such can be cut from a 
double roll. Hence 3 double or 6 single rolls will be needed for 
the ceiling. 

11. The cost of the paper for the ceiling will be 6 x $.45 or 
$2.70. 

12. .*. the total cost will be $ 5.40 + $ 2.70 or $ 8.10. 


EXERCISE 76 

1. How many rolls of paper 18 in. wide will be required 
to paper a room 15 ft. long, 13 ft. wide, and 9 ft. high, no 
allowance being made for openings ? (Buy double rolls.) 

2. How much will it cost to paper a room 14 ft. square 
and 9 ft. high, with paper 18 in. wide, at $.75 per roll? 
Allow 80 sq. ft. for doors and windows. (Include ceiling.) 

3. Plow much will it cost to paper a room 15 ft. long, 
11 ft. wide, and 8 ft. 6 in. high with paper 18 in. wide, 
at $ 1.10 per roll, allowing 175 sq. ft. for doors, windows, and 
baseboards ? 

4. Find the approximate cost of paper for the walls and 
ceiling of the living room of the plan shown in Example 8, 
p. 188, if the walls are to be covered with paper at $ .95 per 
roll and the ceiling at $ .50 per roll. (Assume that the 
ceiling is 9 ft. high, and do not make any allowance for the 
baseboards.) 

5. For the dining room of the same house, paper for the 
side walls is to be purchased at $ .70 per roll and for the 
ceiling at $ .45 per roll. What is the approximate cost ? 


156 


NEW HIGH SCHOOL ARITHMETIC 


6 . How many rolls of paper 18 in. wide will be required 
to paper a room 21 ft. long, 15-J- ft. wide, and 8 ft. 9 in, 
high, allowing for two doors each 3 ft. 6 in. wide and 3 win¬ 
dows each 3 ft. 8 in. wide ? 

7. Eind the approximate cost of papering a room 18 ft. 
long, 14 ft. 6 in. wide, and 9 ft. high with paper 18 in. wide, 
at $ 1.19 per roll. Allow for two doors 3 ft. 6 in. wide, two 
windows 2 ft. 9 in. wide, and a group of windows 8 ft. wide. 

8 . A room 16 ft. long, 12 ft. wide, and 9 ft. high has two 
doors each 3 ft. 8 in. wide, two windows each 3 ft. wide, and 
a window group 7 ft. wide. The doors are 7 ft. high, and 
the windows are 5 ft. 9 in. high. Estimate the number of 
rolls of 18 in. paper required for the walls by using in turn 
each of the methods suggested in § 105. 

106. Carpeting Rooms is less common than formerly since 
most people prefer to purchase rugs. 

Carpeting comes usually 27 in. wide. If there is a dis¬ 
tinct pattern in the carpet, then there is a certain amount 
of waste in cutting to match the pattern. 

Example. If a room is 15 ft. 4 in. long and 11 ft. 9 in. 
wide, how much will the carpet cost to cover it, if the 
carpeting is 27 in. wide and costs $ .63 per yard ? Assume 
that there is a waste of 11 in. on each strip except the first 
in matching the pattern. 

Solution. 1. The natural way to run the strips is lengthwise 
of the room. There will be needed as many strips as 27 in. is con¬ 
tained times in 11 ft. 9 in. or in 141 in. 

.*. 6 strips are needed, as 141 27 = 5 + . 

2. The first strip must be 15 ft. 4 in. long. 

Each of the others must be 11 inches longer, namely 16 ft. 3 in. 
There are 5 such strips. 

3. .*. there is required 15 ft. 4 in. + 5 x (16 ft. 3 in.) or all 
together 96 ft. 7 in. 

4. But 96 ft. 7 in. = 32 yd. 

5. .-. the cost will be 32x $ .63 or $20.28. 


PRACTICAL GEOMETRY AND APPLICATIONS 157 


EXERCISE 77 

1. Determine the cost of carpeting the same room as in 
the preceding example, running the strips crosswise of the 
room. 

2. How many yards of carpeting 27 in. wide are needed 
for a floor 18 ft. long and 14 ft. wide, assuming that there is 
not any waste in cutting? (Run carpet lengthwise.) 

Make a scale drawing of the room and the strips of carpet, 
using the scale \ u — 1 ft. 

3. How many yards of carpet 27 in. wide are needed for 
a floor 20 ft. 10 in. long and 16 ft. 4 in. wide, assuming no 
waste in cutting ? 

4. Straw matting comes one yard wide. What will it 
cost to cover a floor 14 ft. 3 in. wide and 15 ft. 6 in. long 
with matting costing $ .85 per yard ? 

5. Find the cost of carpeting a room 16 ft. 9 in. long 
and 12 ft. 2 in. wide with 27 in. carpet at $ 1.75 per yard. 
Assume a waste of 10 in. on each strip except the first. 

6. How many yards of carpet 27 in. wide will he re¬ 
quired for a floor 16 ft. 9 in. long and 13 ft. 4 in. wide if 
there is a waste of 4^- in. on each strip except the first? 

7. Which way should the strips be run to carpet most 
economically a floor 18 ft. 3 in. by 15 ft. 6 in. with 27 in. 
carpet, assuming no waste in cutting? 

8. Linoleum comes in strips 6 ft. wide or 12 ft. wide. 
It is sold by the square yard. How many square yards of 
linoleum are required for a floor 18 ft. long and 14 ft. 3 in. 
wide, assuming that the floor is covered in the more eco¬ 
nomical manner? 

9. A bathroom floor is 7 ft. wide and 11 ft. long. 
Which width of linoleum will cover it most satisfactorily? 
How much should be purchased and what will be the cost 
at $ 1.85 per square yard? 


158 


NEW HIGH SCHOOL ARITHMETIC 


10. Determine the cost of a single color chenille rug, 9 ft 
by 12 ft. in size, at $ 11.50 per square yard. (Approxi¬ 
mate 1918 price.) 

11. Determine the cost of a single tone rug 14 ft. 8 in. 
wide and 21 ft. 6 in. long at $ 12.25 per square yard. 

12. The adjoining figure is the plan of a kitchen. A is a 
built-in cupboard V X 2'. C is a stationary radiator extend¬ 
ing 18" from the wall. B is a gas 
stove, on legs but not easily moved ; 
it extends 30 in. from the wall. 

D is the sink, open below, without 
legs, extending 2' from the wall. 

The owner decides to purchase 
a piece of linoleum to cover that 
part of the floor that is actually 
used. What size piece of linoleum 
what would be the cost at $ 1.75 per 

107. Area of a Parallelogram.—In Eig. 1 below is a 
parallelogram whose base is b units in length and whose 
altitude is a units. Notice that the parallelogram is divided 
into parts I and II by the altitude HK. 


<-1 

A 

3--X 

D 


B 

E 


would you order, and 
square yard ? 


ii 



b K 

Fig. 1. 


H 



K b 

Fig. 2. 


If such a figure were cut from paper, and then cut on the 
line HK , the part marked I could be placed at the right of 
II as shown in Eig. 2, thus forming a rectangular piece of 
paper. The rectangle and the parallelogram inclose the 
same amount of surface. In both the altitudes are a units 
in length and the bases are b units. 

The area of the rectangle is ab square units. 

















PRACTICAL GEOMETRY AND APPLICATIONS 159 


Hence the area of the parallelogram is ab square units. 
Therefore, in abbreviated form (see § 102), 

Rule. — The area of a parallelogram equals the product of 
its base and altitude. 

The formula for the area of a parallelogram is 
A — ab. 

EXERCISE 78 

1. What is the area of a parallelogram whose: 

a. altitude is 10 ft. and whose base is 13 ft. ? 

b. altitude is 12^ ft. and whose base is 6|- ft. ? 

c. altitude is 5 ft. 4 in. and whose base is 7 ft. 6 in. ? 

d. altitude 2.3 ft. and base 6.75 ft.? 

e. base is 15 rd. and whose altitude is 6 yd. ? 


108. Area of a Triangle. — In Fig. 1 below, the altitude 
AD of the triangle is a units in length and the base is b 
units. 



D b b 

Fig. 1. Fig. 2. 


If two such triangular pieces were cut from paper and 
placed together as shown in Fig. 2, the resulting figure 
would have the form of a parallelogram. The base of the 
parallelogram would be b units in length and the altitude 
a units. Hence the area of the parallelogram would be 
ab square units. Therefore the area of the triangle must 
be \ ab square units. 

R U le. — The area of a triangle is one half the product of its 
base and altitude. 

The formula is A = 








160 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 79 

1. In the figure above, suppose that the altitude of the 
triangle is 4 in. and the base is 10 in. in length. How many 
square units would there be in the parallelogram ? There¬ 
fore, how many must there be in the triangle ? 

2. Answer questions like those in Example 1 when the 
altitude of the triangle is 8 in. and the base is 12 in. 

3. Similarly, if the altitude is 10 in. and the base is 15 in. 

4. What is the area of a triangle whose: 

a. base is 12 ft. and whose altitude is 8 ft. ? 

b. base is 5 yd. and whose altitude is 12 ft. ? 

c. base is 3^ ft. and whose altitude is 2 ^ ft. ? 

d. base is 3.5 ft. and whose altitude is 4.7 ft. ? 

e. base is 12 yd. 1 ft. and whose altitude is 5 yd. 2 ft. ? 

5. What is the base of a triangle whose area is 28 sq. ft. 
and whose altitude is 7 ft. ? 

6. Determine the altitude of a triangle whose area is 
30 sq. ft. and whose base is 6 ft. 

109. A Trapezoid is a quadrilateral having two sides that 
are parallel, as trapezoid ABCD. 

The parallel sides BC and AD are 
the bases and the perpendicular EF 
between them is the altitude. 

In Eig. 1 below, suppose that alti¬ 
tude AE is a units in length, base AD b units, and base BC 
c units. Let G be the middle point of DC. If such a figure 




were cut from paper, the part I could be cut off and placed 
at the right of the part II as shown in Fig. 2. The result- 










PRACTICAL GEOMETRY AND APPLICATIONS 161 


ing figure would be triangular in shape, like triangle ABD. 
The altitude would be a units and the base (b + c) units. 

The area of the triangle would be -J- a(b + c) square units. 

But this is also the area of the trapezoid. Therefore 

Rule. — The area of a trapezoid is one half its altitude mul 
tiplied by the sum of its bases. 

The formula is A = ^ a(b -f c). 

EXERCISE 80 

1. Determine the area of a trapezoid whose: 

a. bases are 10 in. and 8 in. respectively, and whose alti¬ 
tude is 4 in. 

b. bases are 15 ft. and 11 ft. respectively, and whose 
altitude is 6 ft. 

c. bases are 5 yd. and 10 ft. respectively, and whose 
altitude is 2 ft. 

d. bases are 6f yd. and 4 yd. respec¬ 
tively, and whose altitude is yd. ^ 

e. bases are 5 ft. 8 in. and 11 ft. 4 in. re- c 
spectively, and whose altitude is 6 ft. 3 in. 

2. Bind the area of the figure repre¬ 
sented by the adjoining drawing. 

3. The adjoining figure represents the end 
of a barn. If the barn is 35 ft. long, find 
the cost of painting the ends and the vertical 
parts of the sides at 30 4 per square yard. 

4 . Bind the area of the figure below. 

















162 


NEW HIGH SCHOOL ARITHMETIC 


110. The triangle ABC, having a right angle at A, is 
called a Right Triangle. The side BC opposite the right 

angle is called the Hy¬ 
potenuse of the right 
triangle. 

Upon the sides of the 
triangle, squares are 
drawn. These squares 
can he divided into the 
small squares shown 
in the figure. By 
counting the small 
squares, it becomes 
clear that the square 
upon the hypotenuse 
of the triangle equals 
the sum of the squares 
upon the other two sides. Since the area of a square is the 
square of its side, 

Rule. — The square of the hypotenuse of a right triangle 
equals the sum of the squares of the other two sides. 

Example 1. The sides of a right triangle are 7 in. and 
9 in. respectively. Find the length of the hypotenuse. 

Solution. 1. Let h = the length of the hypotenuse. 

2. .*. A 2 = 7 2 + 9 2 = 49 + 81 = 130. 

3. h = Vl30 = 11.4+ 

Since the square of the hypotenuse equals the sum of the 
squares of the other two sides, then the square of one of the 
sides must equal the square of the hypotenuse minus the 
square of the other side. 

Example 2. Find the third side of a right triangle 
whose hypotenuse is 19 in. and whose other side is 12 in. 



















PRACTICAL GEOMETRY AND APPLICATIONS 163 


Solution. 1. Let x — the length of the third side. 

2. Then x 2 = 19 2 - 12 2 = 361 - 144 = 217. 

3. .-. a = V217 = 14.73+ 


EXERCISE 81 


1. Find the hypotenuse of a right triangle when the 
sides are: 

a. 6 in. and 8 in. respectively. 

b. 7 in. and 2 ft. respectively. 

c. 13 ft. and 20 ft. respectively. 

2 . What is the other side of a right triangle whose: 

a. hypotenuse is 35 ft. and one side is 21 ft. ? 

b. hypotenuse is 34 ft. and one side is 17 ft. ? 

c. hypotenuse is 15 yd. and one side is 15 ft. ? 

3. Find the length of the diagonal of a square whose 

sides are 8 in. in length. 

4. Find the length of the diagonal of a rectangle whose 
base is 42 in. and whose altitude is 16 in. 


5. What must be the length of a guy wire to brace g 
telephone pole 35 ft. high if it is attached to a stake 10 ft. 
from the foot of the pole ? 

6. Guy wires are to be attached to the top of a wireless 
aerial 120 ft. high and at points 75 ft. from the foot of the 
aerial. How long must they be, exclusive of the part used 
in making the fastenings ? 

7. How long must a pole be cut to serve as a brace at 

a corner post of a fence, if it is to extend from the top of 
a post 5 ft. high to the bottom of a post distant 12 ft. from 
the first one ? hi 

8 . Determine the 
lengths of the differ¬ 
ent oblique lines in 
the adjoining figure. 









164 


NEW HIGH SCHOOL ARITHMETIC 


111. Computing the Area of a Triangle When the Lengths 
of its Three Sides are Known. — The following rule is proved 
in geometry. 

Rule. — i. From one half the perimeter of the triangle sub¬ 
tract in turn each of the three sides, forming three remainders. 

2 . Multiply together one half the perimeter and the three 
remainders found in Step 1. 

3 . The area of the triangle is the square root of the product 
found in Step 2. 

Example. Find the area of the triangle whose sides are 
13 in., 14 in., and 15 in. respectively. 

Solution. 1. One half the perimeter = \ (13 + 14 + 15) = 21. 

2. 21 - 13 = 8; 21 - 14 = 7; 21 — 15 = 6. 

3. .*. the area = V21 x 8 x 7 x 6 = V7056 = 84. 


EXERCISE 82 


Find the area of a triangle when its sides are: 


1. 9 yd., 15 yd., and 18 yd. respectively. 


2. 9 ft., 10 ft., and 13 ft. respectively. 

3. 13 ft., 13 ft., and 10 ft. respec¬ 
tively. 

4. A railroad cuts across a corner of 
a field, leaving outside lines as indicated 
in the adjoining figure. Determine the 
number of acres in the field. 

5. Find the area of a piece of ground 
having the dimensions indicated in the 
adjoining figure. 


34 rd. 









PRACTICAL GEOMETRY AND APPLICATIONS 165 


112. A Circle is a closed curved line in a plane, every 
point of which is equally distant from a point within called 
the center ; as circle ABD. 

A part of a circle is called an 
Arc; as arc AB. 

Note. — This definition, while 
differing from that given formerly in 
all hooks and still in some elementary 
books, is the accepted one in modern 
mathematics. 

A Radius of a circle is the distance from its center to any 
point of the circle ; as 0A. 

A Diameter of a circle is a straight line through the center 
of the circle, having its ends on the circle; as DOB. 

113. The Circumference of a Circle is the length of the 
circle. (See the note in § 112.) By means of a “ tape 
measure,” it is possible to obtain a straight line whose 
length is approximately equal to that of a circular object. 
If this length be divided by the length of the diameter, the 
quotient will be found to be a little more than 3. 

Note. — A class will find it an interesting experiment to 
measure several circular objects in this manner and determine the 
quotients as outlined above. 

The following rule is proved in geometry. 

Rule. — To find the approximate circumference of a circle, 
multiply its diameter by 3.1416. 

The multiplier 3.1416 is represented by the Greek letter 
t r, read pi. Hence if O represents the circumference, D the 
diameter, and R the radius, then 

C = t tD = 27 rR. 

Another approximate value of nr is 3^. This is suffi¬ 
ciently accurate for most practical purposes. 




166 


NEW HIGH SCHOOL ARITHMETIC 


114. Finding the Area of the Interior of a Circle. — The 

interior of a circle cannot be measured exactly by any unit 
of square measure. 

By drawing radii, the interior of a 
circle can be divided into sixteen or more 
equal parts. Each part is approximately 
triangular in shape, — the base being an 
arc of the circle and the altitude a radius 
of the circle. 

The area of each part then is, approximately, one half 
the radius multiplied by ts- of the circumference. When 
the areas of all sixteen parts are added, thus obtaining the 
area of the circle, then 


Xai 6 

7 2 / \ 

Aim 

/ /■ 

1 

\ 

1 12^/7 


/ 

\ 

V 1 / 9 

sVy 


Rule I. — The area of a circle equals one half its radius 
multiplied by the circumference of the circle. 


The formula is A == } EC. 
Since the circumference equals 2ttE, then 
A == \ E x 2 ir JR = 7 tE 2 . 


Rule II. — The area of a circle equals 7r times the square of 
the radius. 


Note 1. — Observe that these rules, also, are expressed in the 
commonly abbreviated form. (See § 102.) 

Note 2. — If the radius is expressed in inches, the area is ex¬ 
pressed in square inches. 


Example 1. What is the circumference and area of a 
circle whose radius is 13 inches ? 

Solution. 1. The radius = 13. 

2. .-. C = 2 irR = 2 x 3 1 x 13 = 81} in. 

3. Also A = irR 2 = 3} x 13 2 = 531} sq. in. 

Example 2. Find the diameter of a circle whose area 
is 48 sq. ft. 




PRACTICAL GEOMETRY AND APPLICATIONS 167 


Solution. 1. Since the area = ir x R 2 , then divide 48 by 
3.1416 in order to get R 2 ; 

that is, R 2 = 48 - 3.1416 = 15.27+ 

2. .\ R = V1K27 = 3.9+ 

EXERCISE 83 

1. Find the circumference and the area of a circle whose : 

a. radius is 17 in.; b. radius is 3.7 in.; c. diameter is 

25 ft.; d. diameter is 4^ ft. 

2. The diameter of a circle is 45 ft. Express its cir¬ 
cumference in yards. 

3. What is the radius of a circle whose circumference 
is 278.5 ft. ? 

4. Find the radius of a circle whose area is 1 sq. yd. 

5. Over how many square yards of ground can a horse 
graze if it is tied to a stake by a rope 30 ft. long ? 

6. If the diameter of the earth is approximately 7912 
mi., what is the length of the equator ? 

7. How many times must a wheel 34 in. in diameter 
revolve in going one mile ? 

8. What is the circumference of a round iron bar 2.5 in. 
in diameter ? 

9. What is the circumference of a 15-in. pulley wheel ? 

10. What is the total pressure on a piston 16 in. in 
diameter in a steam engine when the pressure is 125 lb. 
per sq. in. ? 

11. The diameter of a Ford piston is 3f in. What is 
the area of its surface? 

12. An acre of ground contains 43,560 sq. ft. How long 
is the radius of a circle whose area is one acre ? What is 
the circumference of the circle ? 


168 


NEW HIGH SCHOOL ARITHMETIC 


13 . Two circles having radii of 7 in. and 10 in. respec 
tively are drawn from the same center. What is the area 
of the ring between them ? 

14 . The number of feet that any point on the rim of a 
grindstone or emery wheel travels in a minute is called the 
surface speed of the stone. About 800 ft. per minute is con¬ 
sidered a good average speed for a grindstone and about 
5000 ft. per minute for an emery wheel. 

a. Find the surface speed of a 9-in. emery wheel revolv¬ 
ing 2100 times per minute. 

b. Find the surface speed of a grindstone 24 in. in di¬ 
ameter revolving at the rate of 75 times per minute. 

15 . Determine the maximum number of revolutions per 
minute at which a 15-in. emery wheel may revolve so that 
its surface speed will not exceed 5000 ft. per minute. 

16 . An emery wheel 15 in. in diameter runs at the rate 
of 1400 revolutions per minute. What is the surface 
speed ? 

17 . What is the rim speed of a fly wheel 16 in. in di¬ 
ameter running 500 revolutions per minute ? 

Solids 

115. Closed Surfaces. — A Spherical Surface is an example 
of a closed surface. 

A simple closed surface incloses a definite part of space, 
separating it from the rest of space. 

A Solid is the definite part of space inclosed by a simple 
closed surface. The surface is called the surface of the 
solid. 

The area of the surface is called the Area of the solid; 
the Volume of the solid is the number of times that the 
solid contains a unit of solid measure, such as the cubic 
inch. 


PRACTICAL GEOMETRY AND APPLICATIONS 169 


116. The solids most commonly encountered are 


(a) 


Rectangular 

Parallelopiped. 


( 6 ) 



Right Circular 
Cylinder. 


Sphere. 




The Rectangular Parallelopiped 


117. The rectangular parallelopiped has six rectangular 
faces. 

A crayon box is a familiar example of a 
rectangular parallelopiped. 

The opposite faces have the same shape 
and area. Any face may be considered 
the Base. The four edges meeting the base 


are perpendicular to the base and are of equal length; 
one of them is the Altitude to that base. 


any 


* f~ 


118. Finding the Volume of a Rectangular Parallelopiped. 

— In the adj oining figure, let AB represent p 

4 in., BD 5 in., and BC 3 in. Imagine 
that the figure is divided into cubic inches. 

There will be four layers of these cubes, 
each containing 3 X 5 or 15 cubic inches. 

Hence the volume of the solid is 4 x 15 
or 60 cubic inches. This suggests the 


I ! 


ITT 


4 ~ 


y\, 
■ \/ x 
A | / 
' i X ' 

/U 


Rule.—The volume of a rectangular parallelopiped equals 
the product of its length, width, and height. 

The formula is F = h x W X H or L WH. 












































































170 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 84 

1. Find the volume of a rectangular parallelopiped if: 

a. its dimensions are 10 ft., 4 ft., and 6 ft. respectively. 

b. its dimensions are 15 in., 7 in., and 9 in. respectively. 

c. its dimensions are 4 yd., 8 ft., and 3 yd. respectively. 

2. Find the number of cubic feet of air space in a school¬ 
room 18 ft. by 23 ft. by 12 ft. 

3. a. Find the number of cubic feet of ice in a piece 
24 in. square and 21 in. thick. 

b. How much does such a piece weigh if a cubic foot of 
ice weighs about 58 lb. ? 

4 . An ice box holds a piece of ice 11 in. wide, 21 in. 
long, and 16 in. thick. How much does such a piece 
weigh ? 

5 . How many tons of ice are there in an ice house 25 
ft. wide, 75 ft. long, and 30 ft. high? 

6. What is the approximate weight of a square bar of 
cast iron 2f in. square and 6 ft. long, if cast iron weighs 
.2616 lb. per cubic inch ? 

7 . What will be the cost of a pile of wood 75 ft. long, 
5 ft. high, and 4 ft. wide at $ 4.75 per cord ? 

8 . How many barrels of water are there in a swimming 
pool 20 ft. by 32 ft. if the water is 5 ft. deep ? 

9. a. How many gallons of water will a tank hold, 
which is 4 ft. by 3 ft. by 2\ ft.? 

b. What will the tank full of water weigh if the tank 
itself weighs 65 lb. and a cubic foot of water weighs 
62-1-lb.? 

10. How many cubic yards of material will be required 
for each 100 ft. of an 18-ft. concrete road if the concrete 
averages 7 in. in thickness ? 


PRACTICAL GEOMETRY AND APPLICATIONS 171 


11 . If a ton of ice occupies about 38 cu. ft., how many 
tons can be put into a small ice house 12 ft. long, 10 ft. 
wide, and 6 ft. deep ? 

12 . If a ton of timothy hay occupies about 500 cu. ft., 
determine the approximate number of tons in a hay pile 
40 ft. long, 22 ft. wide, and 15 ft. high. 

13 . The owners of a greenhouse have a space 24 ft. by 
30 ft. on which they propose to build a coal bin of brick, 
with walls 16 in. thick. How high must they make the 
walls to have a bin that will hold 200 tons ? 

One ton of bituminous coal occupies about 38 cu. ft. 

14 . The water in a ditch flows at the rate of 2 mi. per 
hour. If the ditch is 3 ft. wide and 2 ft. deep, how many 
cubic feet of water pass through the ditch in an hour ? 

15 . How many bushels of small grain can be put into a 
bin 12 ft. long, 6 ft. wide, and 7 ft. high ? 

One bushel occupies about 1^ cu. ft. 

16 . A farmer wants a bin to hold 400 bu. of oats. He 
can make it 8 ft. wide and 10 ft. long. How high must he 
make it? 

17 . How many bushels of potatoes approximately are 
there in a bin 12 ft. long, 5 ft. wide, and 4 ft. high ? 

One bushel occupies about 1| cu. ft. 

18. How many cubic yards of earth must be removed in 
digging a cellar 24 ft. by 30 ft. by 5 ft ? What will this 
cost at 25 1 per cubic yard ? 

19 . What is the average number of cubic feet of air in a 
schoolroom 26 ft. by 32 ft. by 12 ft. if there are 40 pupils 
in the room ? 

20. How many tons of bituminous coal, approximately, 
can a house owner store in his coal room that measures 11 
ft. by 10 ft., if the coal can be piled to an avejuge height 
of 6 ft. ? 


172 


NEW HIGH SCHOOL ARITHMETIC 


119. Concrete Work. — Concrete is made by mixing stone, 
sand, and cement with water. Cement comes in sacks 
which contain approximately 1 cubic foot, — four sacks 
being called a barrel, a. A standard mixture consists of 
1 part cement, 3 parts sand, and 5 parts stone. Since the 
sand and cement fills in spaces between stones, there are 
required for one cubic yard of this mixture 1.16 barrels of 
cement, .52 cu. yd. of sand, and .86 cu. yd. of stone. 

b. Mixed gravel and sand, free from top soil, and not 
containing too much of either sand or gravel, can be used 
in making concrete by mixing 4 sacks of cement to one 
cubic yard of gravel and sand. 

EXERCISE 85 

1. How many bags of cement must a farmer purchase to 
make concrete for a barn floor 20 ft. x 35 ft., if the floor is 
to be 6 in. thick, assuming that he uses the mixture b ? 

2. Estimate the cost of materials for a 7-ft. concrete road 
320 ft. in length, averaging 7 in. in thickness, if gravel can 
be had for $.90 per cubic yard, sand at $.75 per cubic 
yard, and cement at $ 1.65 per barrel, using mixture a. 

3. Determine the cost of the materials for a 4-in. concrete 
floor for a basement 32 ft. long and 27 ft. wide, if the 
concrete is made by mixing 4 bags of cement to one cubic 
yard of gravel. Assume that the gravel costs $1.25 per 
cubic yard and the cement $ 1.85 per barrel. 

4 . Determine the cost of materials for a mile of an 18-ft. 
concrete road 7 in. thick made of the mixture a, figuring 
cement at $ 1.90 per barrel, sand at $ .90 per cubic yard, 
and gravel at $ 1.30 per yard. 

5 . One estimate of the quantity of material required for 
100 sq. ft. of cement sidewalk is 2-1- bbl. of cement, 1^ cu. yd. 
of stone, | cu. yd. of sand, and 2f cu. yd. of cinders. 


PRACTICAL GEOMETRY AND APPLICATIONS 173 


a. What will be required for a sidewalk 600 ft. long and 
5 ft. wide ? 

b. What is the cost of the materials at the prices given 
in Example 4, including cinders at $.75 per cubic yard? 

6. The foundations for a house measure 34 ft. x 28 ft. on 
the outside. What will be the cost of materials for a 
10-inch concrete foundation 8 ft. high, if the concrete is 
made of four bags of cement to one cubic yard of gravel ? 
Eigure the gravel at $ 1.10 per cubic yard, and the cement 
at $ 1.95 per barrel. 

120. Measuring Lumber. — Lumber is measured in terms 
of a unit called a Board Foot. 

A board foot (B. E.) is a piece of wood 1 in. thick and 
1 ft. square. 

Thus, a board 1 in. thick, 1 ft. wide, and 10 ft. long, contains 

10 B. F. 

Clearly, if the hoard were 2 in. thick it would contain 2 x 10 
or 20 B. F. 

If the board were 1 in. thick, 6 in. wide, and 10 ft. long, it is 
equivalent to a board 1 in. thick, 12 in. wide, and 5 ft. long, and 
hence has 5 B. F. 

Rule. —To find the number of board feet in a piece of lumber: 

1. Find its area in square feet. 

2 . Multiply by its thickness in inches; except that, by agree¬ 
ment, all boards less than 1 in. in thickness are measured as if 
1 in. thick. 

Example 1. Find the number of board feet in a 2 in. 
plank, 18 ft. long and 14 in. wide. 

Solution. 1. The area/ of the plank = 18 x If sq. ft. 

2. .*. the number of board feet = 18 x ft x 2 = 42. 

Note. — Observe that the length in feet multiplied by the width 
in inches, divided by 12, and multiplied by the thickness gives the 


174 NEW HIGH SCHOOL ARITHMETIC 


number of board feet. Expressing these computations as in 
Step 2, permits of cancellation. 

Lumber is sold at a certain price per 1000 board feet 
(per M). 


Example 2. Find the cost of 24 planks 20 ft. long, 
14 in. wide, and 2 in. thick at $ 42 per M. 


Solution. 1. The number of B. F. = 24 x 20 x x 2. 

2. .*. the no. of M = 24 x 20 x If x 2 x T ttW 


3. the cost = $42 x 24 x 20 x 14 x 2 = 

12 x 1000 


EXERCISE 86 
Find the number of board feet in : 

1. A board 12' long, 12" wide and J" thick. 

2. A board 10' long, 10" wide and -J" thick. 

3 . A post 10' long, 4" wide and 4" thick. 

4 . A plank 16' long, 14" wide and 2" thick. 

5 . A plank 18' long, 16" wide and 2" thick. 

6. A timber 24' long, 8" wide and 8" thick. 

Find the cost of : 

7. 175 pieces hemlock 2" x 4" x 18', @ $ 33 per M. 

8. 65 pieces hemlock 2" x 12" x 14', @ $ 36 per M. 

9. 26 pieces cypress 2" x 2" x 12', @ $ 58 per M. 

10. 14 pieces cypress 2" x 6" x 16', @ $58 per M. 

11 . 10 pieces oak 1" x 4" x 10', at $ 95 per M. 

12. 15 pieces white pine 2" x 2" x 14', @ $ 85 per M. 

13 . 150 pieces hemlock 1" x 8" x 10', @ $ 37.50 per M. 

14 . 40 cedar posts 4" x 4" x 10', @ $ 33 per M. 

15 . 60 pieces pine 2" x 14" x 16', @ $ 45 per M. 

16. The following bill of material for a hay rack was 
given in a farmer’s journal: 



PRACTICAL GEOMETRY AND APPLICATIONS 175 


Find the total number of B. F. of pine and oak and the 
cost, if $ 75 per M is charged for the oak, and $47.50 per 
M for the pine. 

No. OF 


Pieces 

Size of Stock 

Material 

2 

2" x 8' 1 x 16' 

No. 1 pine 

1 

2" x 4" x 12' 

Oak 

1 

2" x 4" x 14' 

Oak 

2 

2'' x 4" x 10' 

No. 1 pine 

2 

2" x 6" x 14' 

No. 1 pine 

2 

1" X 10" x 16' 

No. 1 pine 

6 

1 " x 6" x 16' 

No. 1 pine 

1 

1-^ 

X 

X 

OO 

No. 1 pine 

4 

1" x 6" x 16' 

No. 1 pine 


The Cylinder 

121. The cylinder most commonly used should be called 
the Right Circular Cylinder to be accurate. Its flat surfaces 
are circular and are equal; they are called the Bases. The 
perpendicular between the bases is the Altitude of the 
cylinder. The area of the curved surface is the Lateral 
Area of the cylinder. 


122. Measuring the Right Circular Cylinder. — If a sheet 
of paper be cut of such size that it can be wrapped around 
the lateral surface of a 
right circular cylinder, 
so that its edges TR 
and US will exactly 
meet at line MN, then 
the area of that sheet 
and of the lateral sur¬ 
face will be the same. Evidently the sheet of paper when 
flattened out will be rectangular, having its altitude TR 
equal to the altitude MN of the cylinder, and its base RS 
equal to the circle bounding the base of the cylinder. 


A 

s' 

*— 
M 

E3 

1 

1 

1 

1 

1 

1 

1 

1 

D 

B 

c 

F'' 

C 











176 NEW HIGH SCHOOL ARITHMETIC 

The area of the sheet of paper is TR x RS. Therefore 
the 

Rule I. — The lateral area of a right circular cylinder equals 
the altitude of the cylinder multiplied by the circumference of 
the base. 

In Solid Geometry, is proved: 

Rule II. — The volume of a right circular cylinder, or of any 
cylinder, equals the area of its base multiplied by its altitude. 

Example. Find the lateral area and the volume of a 
right circular cylinder whose altitude is 9 inches and the 
radius of whose base is 4 inches. 

Solution 1. The circumference of the base = 2x 3.1416 x 4 


= 25.1328. 


.\ the lateral area = 9 x 25.1328 = 226.19+. 
The area of the base = 3.1416 x 16 = 50.2656. 
.*. the volume = 50.2656 x 9 = 452.39+ 


2 . 

3. 

4. 


EXERCISE 87 


1. Find the lateral area and the volume of a right cir¬ 
cular cylinder whose 

a. radius is 3 in. and altitude 10 in. 

b. radius is 8 ft. and altitude 20 ft. 

c. diameter is 16 ft. and altitude is 25 ft. 

2. Asbestos, used for wrapping furnaces and heating 
pipes, comes in rolls 36 in. wide, containing 100 sq. ft. 
How many feet must be ordered to wrap a furnace 56 in. 
in diameter and 6 ft. high ? 

3 . How many feet must be ordered for wrapping a hot 
water tank 18 in. in diameter and 5 ft. high ? 

4 . Approximately, how many gallons of water are there 
in a cylindrical tank 20 in. in diameter and 6 ft. long, neg¬ 
lecting the thickness of the tank ? 



PRACTICAL GEOMETRY AND APPLICATIONS 177 


5 . A galvanized water tank 30 in. high, has a base 
shaped as in the adjoining figure, with the dimensions in¬ 
dicated. How many gallons of 

water will it hold when filled to t*- —30^'—» 

within 3 in. of the top ? 

6. What must be the diameter 
of a cylindrical tank 60 in. long to 
hold 640 gal. ? (Use tt = -3^.) 

7 . A cistern 5 ft. in diameter is 21 ft. deep. How 
many barrels of 31| gallons will it hold ? 

8. How deep must a cistern be to hold 2800 gallons of 
water if the diameter of the cistern is 6 ft. ? 

9. The inside diameter of a cylindrical silo is 15 ft. 
and the height is 32 ft. How many tons of silage will it 
hold if a cubic foot of silage in a silo of that depth weighs 
40 pounds ? 

10. What is the weight of a round marble column 26 in. 
in diameter and 18 ft. long if a cubic foot of marble weighs 
165 pounds ? 

11. How much radiating surface is exposed on 100 linear 
feet of pipe 2 in. in outside diameter, used in a hot water 
heating system in a garage ? 

12. Find the weight of a circular steel plate -J- in. thick, 
and 4 ft. in diameter. (One cubic foot of steel weighs 490 
pounds.) 

13 . What is the approximate weight of a hollow steel 
lawn roller 24 in. in length, 20 in. in diameter, and -J in. 
thick, if steel weighs about .28 lb. per cubic inch ? Neglect 
the weight of the handle. 

14 . What will be the total weight of the lawn roller of 
the previous example, when it is filled with water ? 

15 . What will be the weight of a steel rod 3 in. in di¬ 
ameter and 15 ft. long, if a cubic foot of steel weighs about 
490 lb. ? 






178 


NEW HIGH SCHOOL ARITHMETIC 


16. What is the approximate weight of a steel cylindri¬ 
cal shafting 3J in. in diameter and 20 ft. long ? 

17 . What will be the weight of a cylindrical steel water 
boiler 6 ft. in length and 16 in. in diameter, if made from 
sheet steel weighing 10.1 lb. per square foot ? 

18 . What is the diameter to the nearest inch of a cylin¬ 
drical water tank whose circumference is about 125j- in. ? 

19 . How many gallons of water can be heated in a cylin¬ 
drical tank 20 in. in diameter and 4 ft. high ? 

20 . How many gallons of gasoline can be stored in a 
cylindrical tank 31 in. high and 36 in. in diameter ? 

THE SPHERE 

123. The sphere is inclosed by a surface every point of 
which is equidistant from a point within 
called the Center. The distance from the 
center to the points of the surface is 
the Radius. A straight line segment, 
through the center, having its ends in 
the surface, is a Diameter of the sphere. 

In Solid Geometry are proved : 

Rule I. — The volume of a sphere is | tt times the cube of 
the radius. 

Rule II. — The area of a spherical surface is 4 ir times the 
square of the radius. 

Example. Find the volume of a sphere whose radius is 
6 in.; also the area of its surface. 

Solution. 1. The cube of the radius = 6x6x6 = 216. 

2. .*. the volume = |x-^x 216 = 905| cu. in. 

3. The square of the radius = 6 x 6 = 36. 

4. .*. the surface area = 4 x ^ x 36 = 452f sq. in. 


P 







PRACTICAL GEOMETRY AND APPLICATIONS 179 


EXERCISE 88 

1. Find the volume of the sphere whose radius is: 

a. 7 in. b. 10 in. c. 12 in. 

2. Find the surface area of each of the spheres in 
Example 1. 

3 . Find the volume and the surface area of the sphere 
whose diameter is: 

a. 16 in. b. 15 in. c. 22 in. 

4 . What will be the weight of a hollow steel spherical 
shell 9 in. in diameter and 1 in. thick if made of steel weigh¬ 
ing .28 lb. per cubic inch ? 

5 . What will be the weight of 5000 steel balls (used in 
ball bearings) if they are \ in. in diameter ? 

6. What will it cost to gild a ball 25 inches in diameter, 
at $ 13.50 a square foot ? 

7 . Assuming that the earth is a sphere whose radius is 
about 3856 miles, what is its area and its volume ? 


SUPPLEMENTARY TOPICS 

A. Problems of the Carpentering Contractor 

124. The contractor determines from the drawings of an 
object the kind and quantity of lumber to order. He must 
always make allowances for the difference between nominal 
size of boards and actual size ; thus, a l"x 10" board dressed 
on one edge and one surface will actually measure if" x 9f". 
Also he must make allowance for waste in cutting. Most 
lumber is sold only in even lengths, so that he must order 
material that will cut up well. 

Assume that a contractor is asked to furnish for a factory 
20 sets of shelves as described in the figures below: 


180 


NEW HIGH SCHOOL ARITHMETIC 


Front 


End 


For the ends he will need 40 pieces 1" x 12" finished cypress o’ 
long. He will probably order 20 pieces 1" X 12" cypress 10' long. 

For the shelves and 
tops he will need 100 
pieces 1" X 12" cypress 
about 58 1 " long. He 
will probably order 50 
pieces 1" x 12" x 10' 
long. 

For the shelf rests 
he will need 160 pieces 
1" X 3" x 1' long. He 
will probably order 16 
pieces 1" X 3" x 10'. 

For the diagonal braces, he will need 20 pieces about 6J' long. 
(The length determined by measurement on the drawing.) This 
makes 130 ft. He will probably order 10 pieces 1" x 3" x 14' long. 

For the back, since each piece of 4" ceiling wood shows only 
34" face, he will require as many pieces 4' long as 34" is contained 
times in 60"; that is, 18+ or 19 pieces. For the 20 sets, he will 
therefore need 20 x 19 or 380 pieces, 4' long. This makes 1520 
linear feet. 



His order then takes the following form: 

70 pieces 1" x 12" x 10' dressed cypress 
16 pieces 1" x 3" x 10' dressed cypress 
1520 linear feet 4" pine ceiling wood 
10 pieces 1" x 3" X 14' dressed cypress 


EXERCISE 89 


1. A carpenter is asked to build four 
sashes to the dimensions 3' x 4' 3", using 
2" x 2" cypress. How many linear feet 
of cypress will he order for the four 
windows, if he figures 4 pieces 4' 6" 
long and 2 pieces 3' long for each win¬ 
dow? How much will this cost him 
@ $70 per M ? 

















































~f X „t 


PRACTICAL GEOMETRY AND APPLICATIONS 18l 


2. For each sash will be required 12 panes of 10" x 12" 
glass. What will the glass cost @ $2.70 per box, if a box 
of 10" x 12" glass contains 60 pieces ? 

3 . The owner of a Sca ietf"=r 

house wants a parti- I t--- ii= ^- ii 

, • j p Wall Windows Wall 

tion across one end of 
a basement room to 

form a coal bin. He wall wan 

plans to build it of 

2 X 4 7 S, 8' long, doubling , Partition / ,_ 

them at the center as j - i 

shown, covering them U L 

with 1" x 10" rough 
boards. He decides to ~\ 

make the bin 7 boards |--- 

high. Make out an .j__ 

order for the necessary |--- 

lumber and determine | ~ ~ ~ ~ 

the cost at $ 34.50 j-- 

per M. -is--- >\ 

(Remember, boards come only in even lengths. Determine 
the economical way of ordering the lumber.) 


Scale ya"= I 


Wall Windows 


Wall 


Partition 






4 . Make out an order for the lumber necessary to build 
the fence described in the following drawings. Dressed 8" 
boards are actually only 7^" wide. 

K-| O'-» 

f 

I 


I"x8" 

Dressed 

Boards 


2x4 


2 x 4' 


^Ground 
"r Level 


-100- 



Fence 

& 

o 

o 

'Storage S 

V 

Yard h 




Fence 


Scale 8 A"=ioo' 



























































182 


NEW HIGH SCHOOL ARITHMETIC 


Determine the cost of the material if the 2 x 4 ? s and 
4 x 4 J s cost $ 33.00 per M and the dressed boards $ 38.00 
per M. 

125. Matched Lumber. — The adjoining figure shows an 
end view of two pieces of tongued and grooved matched 
lumber. Finished floors are always 
laid of such lumber; also, wherever it 
is desired to make a tight wall (e.g. in 
a grain bin) matched lumber is used. 

3" flooring is made from wood nominally 1" thick and 3" 
wide. By the time it is cut and smoothed, it is if" thick 
and shows only face. 

Similarly, 4" and 6" flooring show 3\" and 5^" faces, 
respectively. 

When estimating the amount of flooring, determine first 
the number of linear feet required, allowing one-tenth extra 
for waste in cutting. From the number of linear feet, the 
number of board feet can be determined. 

Example. How many board feet of 1" x 6" flooring are 
required for a floor 12' x 15' in size ? 

Solution. 1. Assume that the flooring runs the long way of 
the room. 

2. Since each board shows 5^" face, there will be needed as 
many strips as 5^'' is contained times in 12'. 

3. This is 27f. Hence 28 strips are needed, each 15' long. 

4. .*. 28 x 15 or 420 linear feet are needed. 

5. Adding one tenth for waste, or 42', makes 462 linear feet to 
order. 

6. Since each board is 6" wide, there are 

462 x £ or 231 B. F. 

EXERCISE 90 

1. How many board feet of 1" x 6" flooring are required 
for the floor of an attic measuring 15' x 34', allowing one 
tenth for waste in cutting ? 





PRACTICAL GEOMETRY AND APPLICATIONS 183 


2. Adjoining is the top view of the framework for a 
platform for a camp tent. Assume that the joists are 
2" x 6"; that 2" x 6" pieces are 
nailed across the ends of the joists ; 
that the joists are spaced on 16" 
centers; that the entire frame¬ 
work is covered with 1" x 6" floor¬ 
ing, allowing one tenth additional 
for waste in cutting. Make out 
an order slip for the lumber needed 
and determine the cost if the 2 X 6’s cost $37.00 per M 
and the flooring costs $ 45.00 per M. 

3. The drawings below represent the back of a billboard 
and one set of braces erected at the back of it. Assume 
that a similar set of braces is attached to each of the four 



\* -is-- m 



vertical 4 x 4’s. All other necessary information is in the 
drawings. Make out an order slip for the lumber required 
to erect this signboard. Find the cost if the 4 x 4 J s and 
the 2 x 4 ? s cost $ 35.00 per M and the flooring $ 42.00 
per M. 




















































184 


NEW HIGH SCHOOL ARITHMETIC 


126. Roofs and Roofing. — The figures below give in out 
line form three common styles of roofs. 




Roofs are constructed by nailing rafters in position ; then, 
if the building is to be shingled, shingle strips of 1" x 4" 
rough wood, spaced about 2" apart, are nailed to the rafters ; 
if a tile roof or a roof covered with building paper is to 
be made, the rafters are covered with a tight board roof, 
often made of flooring. 

Shingles are sold by the thousand, — a thousand being 
equal to 1000 shingles 4" wide. They are usually 16" long, 
and, at the thicker end, called the butt, they are about -§" 
thick. Such shingles are packed in bundles, four bundles 
to a thousand. 

Shingles are laid so that 4" to 5" are exposed to the 
weather. 

1000 shingles 4" to the weather will cover 100 sq. ft. of 
surface; 4^-" to the weather, 125 sq. ft. of surface; and 5" 
to the weather, 133 sq. ft. 

EXERCISE 91 

1. If the total surface area of a roof is 600 sq. ft., how 
many thousand shingles will be needed for it, if they are 
laid 4" to the weather? If they are laid 4^" to the 
weather ? 

2. How many bundles of shingles must be ordered for a 
roof having one surface 24' x 10' and another 24' x 18', if 
the shingles are laid 4^" to the weather ? 










PRACTICAL GEOMETRY AND APPLICATIONS 185 


3 . Prom the figures below : 

a. Determine the length of rafter to order for the roof. 

b. Prom the figure determine the number of rafters 
needed, and then from a and b determine the number of 
board feet in them. 



K-46- 


T 

] — 

iL u U-TT= 

^"-End of Rafters^ 

U l 

1 

i 




1 


8' 

T 


Window 


d „ 

g 


I 


<-^6"-> 




i 


1 


t. 

CM 

_1_ 




-11' 6 — ->k£-'6^k-2^> 


c. Determine the number of linear feet of 1" x 4" shingle 
strips needed for the roof, if they are placed about 2" apart. 

d. Determine the number of bundles of shingles required 
to lay the shingles on the roof 4" to the weather. 

e. Determine the number of bundles of shingles needed 
to cover the walls with shingles laid 5 " to the weather. 

f Determine the number of board feet of 1" X 8" boards 
for the walls, allowing 25% for waste in cutting. 

4. The shed roof of a chicken coop measures 24' X 14' 4". 
What will be the cost of materials for a shingle roof made 
as follows : 

a. Rafters, spaced S’ apart from center to center, at 
$ 33.00 per M. 

b. Shingle strips, spaced 2" apart, at $ 33.00 per M. 

c. Shingles, laid 5" to the weather, at $ 6.50 per M. 

5. What will it cost for materials for the roof of Ex¬ 
ample 4 if it is made as follows : 

a. Rafters as in Example 4. 

b. A tight board roof made of 1" X 6” flooring at $ 42.00 
per M. 



















186 J / NEW HIGH SCHOOL ARITHMETIC 


c. Three-ply roofing paper at $ 3.00 per roll. Such 
paper is 3 feet wide and conies in a roll about 36 feet long. 

6. The drawings on p. 187 show the framework of a 
garage with a gable roof. Find in the drawings : 

a. The sill, made of 2 X 4’s laid on a broad face. 

b. The studding, made of 2 x 4’s, doubled at corners and 
openings of large size, spaced 3' apart, from center to 
center, excepting the corner posts. 

c. The plate, doubled 2 x 4’s. 

d. The ridge board. 

e. The rafters, nailed to the plate and to the ridge board, 
extending beyond the plate and spaced 3' apart, center to 
center, excepting the two end rafters. 

/. The header over the doorway, doubled 2 x 4’s. 

g. The shingle strips, 1" x 4", 2" apart, one at the ridge 
on each side of the ridge, and one at the lower end of the 
roof. 

h. Notice the rise of the gable roof, the distance from 
the top of the plate to the top of the ridge. 

i. Notice the span of the roof, the width of the building. 

j. Notice that the pitch of the roof is the rise divided 
by the span. 

Then answer the following questions and solve the 
examples. 

a '. What would be the rise of a gable roof if the span is 
30 ft. and the pitch is -J- ? \ ? 4 ? 

b'. What is the pitch of a gable roof if the span is 20' 
and the rise is 5' ? 8' ? 10' ? 

c'. Determine the length of rafter for the garage pictured, 
allowing 2 ft. for overhang. 

d'. Since 2 x 4’s come only in even lengths, what length 
must be ordered for the rafters, remembering that about 3 f! 
must be wasted at each end in making the slanting cut ? 


Rise 


PRACTICAL GEOMETRY AND APPLICATIONS 18 ? 


e Make out a memorandum for the lumber required 
for the frame of the garage, as follows: 

Lumber Required for Framing of Garage 


Purpose Number Kind Quantity 

Sills 

Studding 
Plate 
Rafters 
Shingle Strips 
Ridge 

Horizontal braces 
Headers 
Diagonal braces 

f. Suppose that the outside of the building is first cov¬ 
ered with 1" x 8" sheathing, then with paper, and finally 
with siding. Determine the number of board feet of sheath¬ 
ing, not making any allowance for the doors or windows, 
allowing 20% for waste in cutting. 


2 2 x 4 12' 16 B. F. 

2 ? 



Determine also the number of rolls of building paper 
needed, if it comes in rolls of 500 sq. ft. Also determine 
the number of linear feet of 8" cypress siding needed if it 
is laid so that 7" is exposed to the weather. Do not allow 
for the doors or the windows ; add £ for waste. 





























































188 


NEW HIGH SCHOOL ARITHMETIC 


7 . The adjoining figure is a skeleton view of the end of a 
gambrel-roofed barn. Suppose that the barn is 45 ft. long. 

Determine the cost of shingles 
for the slanting surfaces if they are 
laid 4^-" to the weather and cost 
$ 6.25 per M, assuming that the 
roof does not project over the ends 
of the barn. Determine the cost at 
$ 60.00 per M of 12" pine boards 
for the walls, put on vertically. 

8. a. From the floor plans below determine the number of 
linear feet of 3" oak flooring (2J" face) required for the 
living room, halls, and dining room; also the number of 
linear feet of 3" maple flooring for the kitchen. Allow one 
tenth additional for waste in cutting. 




b. Find the cost of the flooring if the oak comes at 
$ 85.00 per M, and the maple at $ 72.00 per M. 

c. Find the cost of a tile floor for the porch at $ .60 per 
square foot. Do not allow for the fireplace chimney or the 
places occupied by the posts. 


















































PRACTICAL GEOMETRY AND APPLICATIONS 189 


9 . Determine the cost of painting the four walls of the 
shed shown in Example 3, at $ .25 per square yard. 

10 . Determine the cost of painting the vertical walls of 
the barn shown in Example 7, at $ .32 per square yard. 


B. Mensuration of Less Common Solids 

127. A Polyhedron is a solid bounded by portions of 
planes called Faces of the polyhedron. The faces intersect 
in straight lines, called the Edges of the polyhedron; the 
edges intersect in points called the Vertices of the poly¬ 
hedron. The straight lines joining any two vertices which 
do not lie in the same face are called the Diagonals of the 
polyhedron. 

128. A Prism is a polyhedron, two of whose faces lie in 
parallel planes, and whose remaining faces, in order, inter¬ 
sect in parallel lines. The parallel faces are the Bases; the 
other faces are the Lateral Faces; the edges which are not 
sides of the bases are the Lateral Edges; the perpendicular 
between the bases is the Altitude; the sum of the areas of 
the lateral faces is the Lateral Area. 


129. A Right Prism is a prism whose lateral edges are 
perpendicular to its bases. 

The lateral faces are evidently rectangular, and the 
lateral edges equal the altitude. 

130. Finding the Lateral Area of a Prism. — Consider the 
face ABED. It is bounded by a rectangle 
whose base is AB and altitude BE, — the same 
as the altitude of the prism. The area of 
rectangle ABED = AB x BE. In like man¬ 
ner, the area of each lateral face equals a side 
of the base multiplied by the altitude of the 
prism. 










290 


NEW HIGH SCHOOL ARITHMETIC 


Therefore 

Rule I. — The lateral area of a right prism is equal to the 
sum of the sides of the base multiplied by the altitude of the 
prism. 

Just as the volume of a rectangular parallelopiped is the 
product of its base by its altitude, so 

Rule II. — The volume of any prism equals the area of its 
base multiplied by its altitude. 


131. The adjoining figure represents a Pyramid. It is a 
polyhedron having one Base bounded by a polygon, face 
ABCDEF, and as many triangular faces o 

with a common vertex, 0, as there are 
sides of the base. The common point 
of the triangular faces is called the Vertex 
of the pyramid; the triangular faces are 
the Lateral Faces; and the sum of their 
areas is the Lateral Area; the triangular A 
faces meet on the Lateral Edges, as AO, 

BO, etc.; the perpendicular from the vertex to the base is 
the altitude, as OG. 



132. A Regular Pyramid is a pyramid whose base is in¬ 
closed by a polygon whose sides are all equal, and whose 
angles are all equal, and whose vertex 
lies on a line perpendicular to the plane 
of its base at the center of the base. 

Note. — A polygon having equal sides 
and equal angles is called a regular polygon. 

Such a polygon has a center. 

It is proved in solid geometry that 
all the lateral edges of a regular pyra¬ 
mid are equal, and that the lateral faces have equal alti 
tudes. 







PRACTICAL GEOMETRY AND APPLICATIONS 191 


The common length of the altitudes of the lateral faces 
is called the Slant Height of the Regular Pyramid, as line OH. 

Note. — Only a regular pyramid has a slant height. 

The line representing the slant height bisects the side of 
the base to which it is drawn. 

133. Finding the Lateral Area of a Regular Pyramid. 

The area of each of the lateral faces is one half its alti¬ 
tude multiplied by its base. But this is one half the slant 
height of the pyramid multiplied by one of the sides of the 
base. When the areas of all the faces are added, there 
results the 

Rule. — The lateral area of a regular pyramid equals one 
half its slant height multiplied by the perimeter of its base. 

Example. Bind the lateral area of a regular pyramid 
having 8 faces, if one side of the base is 6 ft. and the slant 
height is 10 ft. 

Solution. 1. The perimeter of the base is 8 x 6 ft. or 48 ft. 

2. .\ the lateral area = | X 10x 48 = 240 sq. ft. 

134. A Frustum of a Pyramid is that part of a pyramid 
inclosed between the base and a plane parallel to the base. 

The base and the section of the plane 
parallel to the base are called the Bases 
of the frustum; the perpendicular be¬ 
tween the bases is called the Altitude of 
the frustum. 

The lateral faces are bounded by 
trapezoids. If the frustum is the frustum of a regular 
pyramid, the altitudes of the lateral faces are all equal, and 
their common length is called the slant height of the frus¬ 
tum ; as IIII'. 

Note. — Only frustums of regular pyramids have a slant 
height. 






192 


NEW HIGH SCHOOL ARITHMETIC 


The line representing the slant height of a frustum of a 
regular pyramid joins the mid-points of the bases of the 
face on which it is drawn. 

135. Finding the Lateral Area of a Frustum of a Regular 
Pyramid. — In the figure of § 134, the area of trapezoid 
ABB'A' is one half its altitude multiplied by the sum of its 
bases; but this is one half the slant height of the frustum 
multiplied by the sum of one side of the lower base and 
one side of the upper base. When the areas of all the 
lateral faces are found in this manner and added, there 
results the 

Rule.—The lateral area of a frustum of a regular pyramid 
equals one half its slant height multiplied by the sum of the 
perimeters of its bases. 

136. The following rules for the volumes are proved in 
solid geometry. 

Rule I. — The volume of any pyramid equals one third its 
altitude multiplied by the area of its base. 

Rule II. — The volume of a frustum of any pyramid equals 
one third its altitude multiplied by the sum of the areas of its 
lower base and its upper base, plus the square root of the product 
of these areas. 

Example. Find the volume of a frustum of a regular 
pyramid whose lower base is a square having a side of 8 in., 
whose upper base is a square having a side of 5 in., and 
whose altitude is 6 in. 

Solution. 1. The area of the lower base is 8 x 8 or 64 sq. in. 

The area of the upper base is 5 x 5 or 25 sq. in, 

2. The volume = |x 6 x (64 + 25 + V64 x 25) 

= I x 6 x (64 + 25 + 8 x 5)= 2 x 129 

= 258 cu. in. 



PRACTICAL GEOMETRY AND APPLICATIONS 193 


EXERCISE 92 

1. Find the lateral area and volume of a right prism 
whose altitude is 14 in., having for its base a right triangle 
whose sides are 5 in., 12 in., and 13 in. 

2 . Find the lateral area of a regular pyramid whose 
base is a square 5 ft. on a side, and slant height 13 ft. 

3 . What is the volume of a pyramid whose altitude is 
24 in., having for its base a right triangle whose sides are 
8 in., 15 in., and 17 in. ? 

4 . Find the lateral area of a frustum of a regular pyra¬ 
mid whose lower base is a square 10 ft. on a side, upper 
base a square 5 ft. on a side, and whose slant height is 
12 ft. 

5. The volume of a pyramid, whose base is a square, is 
1089 cu. in., and its altitude is 27 in. Find the length of 
each side of the base. 

6. Find the volume of a frustum of a pyramid whose 
lower base is a rectangle 10 in. by 6 in., upper base a 
rectangle 5 in. by 3 in., and whose altitude is 17 in. 

7. The lateral area of a regular pyramid is 1554 sq. in. 
The base is a triangle whose sides are all equal, and the 
slant height is 37 in. Find the length of each side of the 
base. 

8. The lateral area of a frustum of a regular pyramid 
is 936 sq. in. The lower base is a square 18 in. on a side, 
and the upper base is one 6 in. on a side. Find the slant 
height of the frustum. 

9. A monument is in the form of a frustum of a square 
pyramid 8 ft. in height, surmounted by a square pyramid 
2 ft. in height. If each side of the lowet base of the 
frustum is 3 ft., and each side of the upper base is 2 ft., 
find the volume of the monument. 


194 


NEW HIGH SCHOOL ARITHMETIC 


10. The volume of a frustum of a pyramid is 210 cu. in. 
The lower base is a right triangle whose sides are 6 in. and 
8 in., and the upper base is a right triangle whose sides are 
3 in. and 4 in. Eind the altitude of the frustum. 

11. The base of a square pyramid is 14 in. on a side, and 
the altitude is 24 in. Find its lateral area and volume. 

12. The volume of a rectangular parallelopiped whose 
base is a square is 896 cu. ft., and its altitude is 14 ft. Find 
the length of each side of its base, and its lateral area. 

13 . The volume of a box is 83.25 cu. ft., and the dimen¬ 
sions of its base are 5 ft. and 4.5 ft. Find its height, and 
the area of its total surface. 

14 . The volume of a cube is 4-Li- cu. ft. What is the 
area of its entire surface in square inches ? (Cube root 
needed.) 

15 . A monument whose height is 12 ft. is in the form of 
a pyramid with a square base, 2 ft. 10 in. on a side. Find 
its weight if a cubic foot of the stone weighs 180 lb. 


CONES 

137. Right Circular Cone. — In the adjoining figure, let 
ADB represent a circle with center C. Let OC be a line 
perpendicular to Base ADB at C. Assume 
that line OA moves so that point 0 is 
stationary and point A travels on circle 
ADB. This line forms a surface called a 
Conical Surface. 

The solid bounded by the base ADB and 
the conical surface is called a Right Circular 
Cone. 0 is the Vertex of the cone. Line A 
OC is the axis of the cone, and also the Alti¬ 
tude ; the conical surface is the Lateral Surface; the line from 
0 to any point of the circle bounding the base is the Slant 
Height as OA. 




PRACTICAL GEOMETRY AND APPLICATIONS 195 


Note. — In solid geometry, other cones as well as right circular 
cones are studied. 

138. Measuring the Right Circular Cone.—In solid geom¬ 
etry the following rules are proved: 

Rule I. — The lateral area of a right circular cone equals the 
circumference of its base multiplied by one half its slant height. 

Rule II. — The volume of a right circular cone, in fact of 
any cone, equals the area of its base multiplied by one third its 
altitude. 

139. A Frustum of a cone is that part of the cone included 
between the base and a plane parallel to the base. 

The base of the cone and the section of 
the cutting plane parallel to it are the 
Bases of the frustum; the perpendicular 
between the bases is the Altitude, as line 
EF\ the part of the slant height of the 
cone included between the bases, line AC, 
is the Slant Height of the frustum of the cone. 

In solid geometry are proved : 

Rule i. — The lateral area of a frustum of a right circular 
cone equals the sum of the circumferences of the bases multi¬ 
plied by one half the slant height. 

Rule II. _The volume of a frustum of a right circular cone, 

or of any cone, equals the sum of the areas of the bases plus the 
square root of the product of these areas, multiplied by one 
third the altitude of the frustum. 

Example. Eind the volume and the lateral area of a 
frustum of a right circular cone whose slant height is 8 in., 
whose upper base diameter is 6 in., and whose lower base 
diameter is 14 in. 

Solution. 1. The circumference of the lower base = 14 tt in , 
and of the upper base = 6 tt in. 










196 


NEW HIGH SCHOOL ARITHMETIC 


2. .. the lateral area = \ x 8 x (14 v + Qtt) = 4 x 20 7r — 80 7r in 

3 . = 80 x 3.1416 = 251.328 = 251.3+ sq. in. 

4. The area of the lower base = 7 2 it sq. in. = 49 7 r sq. in. 

The area of the upper base = 3 2 it sq. in. = 9 tt sq. in. 

5. In the figure of § 139, the altitude = EF or AD. 

6. CD = CF-AE = 7-3=4. 

7. In right triangle A CD, AZ> 2 = 8 2 — 4 2 = 48. 

AD = V48 = 6.9+ in. 

8. the volume = | X 6.9 x (49 tt + 9 tt + V49 tt x 9 tt) 

= 2.3 x (49 tt + 9 7r + 7 7r x 3) 

= 2.3 x 79 7 t 
= 181.7+ cn. in. 


EXERCISE 93 

1. Eind the lateral area and the volume of a right circu¬ 
lar cone whose altitude is 18 in. and the radius of whose 
base is 6 in. 

2. A tent, in the shape of a right circular cone, has a 
slant height of 20 ft. and a diameter at its base of 30 ft. 
How many square yards of material were used in construct¬ 
ing it ? 

3 . Eind the volume of a frustum of a right circular cone 
whose altitude is 5 ft. and the diameters of whose bases are 
30 in. and 40 in. respectively. 

4 . How many cubic feet are there in a column whose 
length is 24 ft., the diameter of whose larger end is 14 in., 
and the diameter of whose smaller end is 10 in. ? 

5 . What is the volume of a frustum of a right circular 
cone whose altitude is 13 in. and the radii of whose bases are 
7 in. and 3 in. respectively ? 

6 . Eind the lateral area of a frustum of a right circular 
cone whose slant height is 15 ft. and the radii of whose 
bases are 5 ft. and 3 ft. respectively. 



PRACTICAL GEOMETRY AND APPLICATIONS 197 


7 . The lateral area of a right circular cone is 188.496 
sq. in. and the radius of the base is 6 in. Find the slant 
height, the altitude, and the volume. 

8. The volume of a frustum of a right circular cone is 
779.1168 cu. ft. and the radii of its bases are 10 ft. and 2 ft. 
respectively. What is the altitude ? 

9 . The altitude of a cone is 12 in., and the radius of its 
base is 7 in. Find the altitude of a cylinder of equal vol¬ 
ume, the diameter of whose base is 10 in. 

10 . The altitude of a frustum of a right circular cone is 
6 ft., and the radii of its bases are 3 ft. and 2 ft. respec¬ 
tively. What is the diameter of a sphere of equal volume ? 

11 . The volume of a right circular cone is 2412.7488 cu. 
in. and its altitude is 12 in. Find the radius of its base. 

12 . Find the radius of a sphere whose surface equals in 
area the lateral surface area of a right circular cone, whose 
altitude is 8 ft. and the radius of whose base is 4 ft. 

13 . The area of a sphere is 314.16 sq. in. What is its 
diameter and volume ? 

14 . How many yards of canvas are needed for a show 
tent having the form of a right circular cone, if the diameter 
of the tent is 60 ft., the height of the vertical wall 10 ft., and 
the height of the highest point of the tent 25 ft. ? 

15 . In Example 14 what is the cubic contents of the tent ? 

16 . What is the volume of an ice-cream cone which is 2 in. 
in diameter and 5 in. long ? 

17 . A Christmas tree ornament has the form of a right 
circular cone. If it is 1.5 in. in diameter and 4 in. long, 
about how many such ornaments can be filled from a box 
containing 2 lb. of candy, if the box is 9 in. long, 5 in. wide, 
and 2.5 in. high ? 


198 


NEW HIGH SCHOOL ARITHMETIC 


C. Graphical Representation 

140. Numerical Relations are often conveyed to the mind 
advantageously by a Graph. 

Example. The figure below represents the average 
retail price in cents per pound of sirloin steak in certain 
years. 


Cents 

0 10 20 30 40 50 60 


1913 












| 


1 

1 

































1925 
















































1929 
















































1930 
















































1931 

















































The numbers above the graph give the Scale. The side 
of a large square represents 10 $ ; of a small square, 2 /• 

The Bar opposite 1913 represents a little over 26 

The Bar opposite 1929 represents a little over 52 ^ and 
less than 53 0, or about 52.5 ^; so one pound of steak in 
1929 cost about 52.5 

Questions 

1. In what year was the cost of sirloin steak greatest ? 

2 . Was the cost in 1931 more or less than the cost in 
1925 ? Than the cost in 1913 ? 

3 . What was the cost of one pound in each year ? 

141. Selecting the Unit when drawing a graph depends 
upon the size of the number to be represented and upon the 
size of the paper upon which the graph is to be drawn. 

In § 140, the person who drew the graph knew that the 
cost of sirloin steak in 1929 was about 52.5 per pound. 
He thought: “If I let a side of one. large square represent 
5/, it will take over 10 such sides to represent 52.5^. 
This would be too wide for the page.” He decided to let a 
side of one large square represent 10^. 





































PRACTICAL GEOMETRY AND APPLICATIONS 199 


142. Another form of the Bar Graph is shown below. 

This graph shows the number of pupils in each of the 
four classes of a certain four-year high school in a recent 
year, the number of pupils in each class who were boys, and 
the number who were girls. 


0 50 100 150 200 250 


- 


' 





■ 






"1 

[ 




















(//A 












SSfiJs 


... 




1 










Boys 


I i Girls 


Thus, in the ninth-grade class there were 275 pupils. Of these 
pupils, 150 were boys and the remainder were girls. 

Questions 

1. How many pupils were there in each of the four classes 
of the school ? 

2. Without finding the number of boys and the number 
of girls in each class, answer the following question: 

Was the number of girls in any class one-half the total 
number of pupils in the class ? 

3 . How many boys were there in each class ? 

4 . Determine the number of girls in each class. 

5 . Ask your teacher to appoint a committee of one from 
your class to obtain from the principal of your school the 
number of boys and of girls in each class in your school. 
Then all the pupils in your class may draw a graph like the 
one above for the enrollment in your school. 

6 . Ask your teacher to tell you how many pupils in your 
grade received the marks A, B, C, and D during the last 
marking period. (If you do not use the marks A, B, C, D, 
substitute whatever marks are used in your school.) 

Draw a single bar graph to represent the total number of 
pupils in your class. Then divide it into pieces which will 
represent the number of pupils who received each of the 
different marks. 






































200 NEW HIGH SCHOOL ARITHMETIC 
143. A Circular Distribution Graph appears below. 


This graph represents the popu¬ 
lation of the world in a recent year 
and the part of it living on each 
of the continents. 


Questions 

1. How did the population of Asia compare with that of 
the whole world ? 

2. About what part of the population of Asia was the 
population of Europe? 

3 . Leaving out the population of Asia, how did the popu¬ 
lation of Europe compare with the population of all the 
other continents? 



4 . What two continents appear to have about the same 
population ? 

5 . How does the population of South America appear to 
compare with that of North 
America ? 

6 . The adjoining graph repre¬ 
sents the total amount of artifi¬ 
cial gas sold for various purposes 
in Chicago in a recent year. 

a. For what purpose was the 
largest part of the gas sold ? 

b. For what purpose was the 
least amount of gas sold ? 

c. For what two purposes were about equal amounts sold? 

d. About what part of the whole amount was used in 
industry ? 




PRACTICAL GEOMETRY AND APPLICATIONS 201 


144. A Broken-line Graph appears below. 

The total immigration into the United States during 
recent years is given in the following table. 


Year 

Immigrants 

Year 

Immigrants 

1925 

294,314 

1928 

307,255 

1926 

304,488 

1929 

279,678 

1927 

335,175 

1930 

241,700 


To represent these numbers graphically, we Round Off 
each to hundred thousands. Thus: 


In Year 

1925 

1926 

1927 

1928 

1929 

1930 

The immigration in 







100,000’s was 

2.9 

3.0 

3.4 

3.1 

2.8 

2.4 


3 

ii 

'S S 
-S jj 

c 1-1 
p 

W 1 


This graph shows that the immigration was increasing 
from 1925 to 1927, and that it has been decreasing since 
then; that the increase during 1926 was greater than the 
increase during 1925; that the decrease since 1927 has been 
at a regular rate. 

The graph is a broken line (§ 97). 













































































202 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 94 

1. Graph, of the number of pupils iu school in the United 
States during each of certain years. 


1920 

| | | | 


























1922 


























1924 

1926 

1928 


till 
























































0 6 10 15 20 

Million Pupils 


Thus, in 1920, there were about 16.2 million or 16,200,000 pupils 
in the schools. 

a. How many pupils were in the schools in 1922 ? 

b. About how many more pupils were in the schools in 
1924 than in 1922 ? In 1928 than in 1926 ? 

c. What kind of graph is this ? 

2 . Graph showing the values of cotton and wheat crops 
in each of certain years. 



a. In what year were the values of cotton and wheat 
crops about equal ? 

b. Which crop usually has the larger value ? 

c. In which year was the value of the cotton crop 
greatest ? 

d. How much greater than the value of the wheat crop 
was the value of the cotton crop in 1929 ? 














































































PRACTICAL GEOMETRY AND APPLICATIONS 203 


3 . A graph, showing the growth in population of Chicago 
and of Philadelphia since 1870. 



a. How much more than the population of Chicago in 
1870 was the population of Philadelphia? 

b. In what year were the populations equal ? 

c. In which city is the population increasing faster ? 

4 . Graph showing the number of games won and lost by 
the teams of the American League in a recent year. 


Team 


Phila. 

N. Y. 



































3 

Wash. 

Cleve. 

St. Louis 

Boston 



































T 



































J 



































I 















i 




















J 


Detroit 

Chicago 















4- 





















r 

1 

3 



2 

0 




4 

0 

_1- 

6 

0 


8 

0 




1( 

)0 120 140 


H Games won 1 I Games lost 

a. What is the greatest number of games played by any 
team? 

b. What teams lost more games than they won ? 

c. What team won about one-half its games ? 





























































































































204 


NEW HIGH SCHOOL ARITHMETIC 


5 . Draw a graph like that for Example 4 to exhibit the 
games played, won, and lost, by the teams of the National 
League, as given in the following table. 


Team 

Won 

Lost 

Team 

Won 

Lost 

St. Louis . . . 

101 

53 

Pittsburgh . . 

75 

79 

New York . . 

87 

65 

Philadelphia . . 

66 

88 

Chicago . . . 

84 

70 

Boston . . . 

64 

90 

Brooklyn . . . 

79 

73 

Cincinnati . . 

58 

96 


6 . a. Draw a broken-line graph to show the enrollment 
in one high school during the last six years. 


Year 

Enrollment 

Year 

Enrollment 

6 years ago . . 

415 

3 years ago . . 

592 

5 years ago . . 

475 

2 years ago . . 

620 

4 years ago . . 

540 

1 year ago . . 

685 


Suggestion. Round off the result to hundreds and tenths of a 
hundred. For example, 415 = 4.2 hundred. 

b. If possible, secure the enrollment in your school dur¬ 
ing these same years. On the same piece of graph paper, 
draw the graph to represent the enrollment in your school. 

7 . The following table gives the total number of automo¬ 
biles produced in the world in certain years. 


Year 

Number 

Year 

Number 

1927 


4,158,966 

1929 

6,277,451 

1928 


5,203,139 

1930 

4,109,231 


a. Express each of the numbers in millions , correct to 
tenths of a million. 

b. Draw a bar graph to represent the results in part a. 

































PRACTICAL GEOMETRY AND APPLICATIONS 205 


8 . The six highest buildings in New York City at the 
present time, and their heights, are : 


Empire State Building 1250 ft. Wool worth Building . 792 ft. 

Chrysler Building . . 1046 ft. City Bank. 745 ft. 

Bank of Manhattan . 925 ft. 35 Wall Street ... 715 ft. 


Represent these heights by a vertical bar graph. 










24 

24 

-t 

24 W 







Tt* 

CM 





24 24 24 


BASE LINE E 


D. United States Government Land Survey 

145. The land area of the Central and Western States 
was surveyed by the United States Government. 

Surveyors first marked the true meridian of the earth 
(running north and south) through some prominent point 
in a district. This is called 
the Principal Meridian. 

The surveyors also marked 
the parallel of latitude running 
(east and west) through this 
point. This parallel is called 
a Base Line. They also marked 
the parallels of latitude at in¬ 
tervals of twenty-four miles 
north and south of the base 
line. 

On each of these parallels, points were established at inter¬ 
vals of twenty-four miles east and west of the principal 
meridian. Through these points, meridians were marked 
between the parallels, as shown in the figure adjoining. 
Because of the curvature of the earth, these meridians are 
a little less than twenty-four miles apart at the top, although 
they are full twenty-four miles apart at the bottom. 

As a result, the land was divided into tracts which are 
approximately twenty-four miles square. (See also Big. 2.) 

Each of these tracts was divided by true parallels of lati¬ 
tude and true meridians into parts approximately six miles 


Fig. l. 



















206 


NEW HIGH SCHOOL ARITHMETIC 


square. These parts are called Townships. They were 
meant to be 6 mi. square. The east and west boundaries 
of a township are full six miles in 
length; also the south boundary of 
each township of the bottom row of 
each 24 mi. square tract is also six 
miles in length; but the south bound¬ 
aries of all the others and the north 
boundaries of all the townships are 
less than six miles in length. 

The townships in a line north and 
south form a Range; those in a line east and west form 
a Tier. 

The ranges are numbered east or west from the principal 
meridian; the tiers are numbered north or south from the 

baSe Une - TOWNSHIP 

Thus, in Fig. 2 , if A M be assumed N 

to be the base line and MT the Third _ 5 4 3 21 

Principal Meridian, then township 7 8 9 io u 12 m 

I is 3 N., Range 2 W. of the Third is 17 16 15 14 ~\3 ^ 

Principal Meridian. W ^ ^ ^ | E 

- <5 

Each township is divided into 30 29 28 27 25 25 

thirty-six parts that are approxi- 31 32 33 34 35 36 

mately one mile square, called S|X M ' LE s 

Sections. These parts are num- S 

bered as shown in the adjoining 

figure, Fig. 3. Actually, because 
of the fact that most of the 
townships are not six miles square, 
and because of the manner of sub¬ 
dividing the townships, the sections 
in the western range and the 
northern tier are slightly less than 
one mile square. 

The adjoining figure represents a 


N 


IV 

V I 

II 

III 


S 


Fig. 4. 


TOWNSHIP 


N 


6 

5 

4 

3 

2 

1 

7 

8 

9 

10 

11 

12 

18 

17 

16 

15 

14 

13 

19 

20 

21 

22 

23 

24 

30 

29 

28 

27 

26 

25 

31 

32 

33 

34 

35 

36 


SIX MILES 
S 


Fig. 3. 


II 






I 











A 6mi. 6mi. 6mi. 6mi. M 

Fig. 2. 

























PRACTICAL GEOMETRY AND APPLICATIONS 207 


subdivision of a section. Field IY would be described as 
the north | of the section; field II, as the S. E. \ of the 
section; field III, as the S. \ of the S. W. \ of the section; 
field I, as the N. E. i of the S. W. \ of the section. 


EXERCISE 95 

1. Since a section is approximately one square mile, how 
many acres does it contain ? 

2. How many acres are there in each of the fields of 
Fig. 4, p. 206 ? 

3 . Describe the location of field Y 
in Fig. 4, p. 206. 

4 . The adjoining figure represents 
Section 22 of Township 3 N., Range w 
23 E. of the Fourth Principal Me¬ 
ridian. Describe the location of each 
of the farms shown in the figure. 

How large is each ? 

5 . Represent by a drawing a farm described as the W. 
J of the N. E. \ of a certain section. 

6 . Represent by a figure a farm that is described as the 
S. W. \ of the N. W. \ of a certain section. 

7 . How many rods of fencing are required to inclose the 
farms described in Example 5 and Example 6 ? 

8. Find the value of the farm in Example 5 at $ 135 per 
acre. 

9. Find the value of the farm in Example 6 at $ 175 per 
acre. 



II 

I 

III 

IV 

V 


VI 









VII. PERCENTAGE 


146. The expression Per Cent is an abbreviation for per 
centum, meaning “ in ” or “ by a hundred.” 

Thus, 3 per cent means 3 in every hundred. 

The symbol for per cent is % ; thus, 3 per cent is written 

3 %. 

Since 3 % means 3 in every hundred, then 3 % of 200 
must be 6, 3 % of 500 must be 15, etc. Evidently 3 % of 
a number is the same as T -| T or .03 of the number. 

147. The Rate Per Cent is the number of hundredths to 
be found. 

The Base is the number of which a certain per cent is to 
be found. 

The Percentage is the result obtained by taking a certain 
per cent of a base. 

148. A Rate Per Cent is usually Expressed in Decimal Form 
when it is used. 

Thus, 27 % = T Vo = .27. 

5.3%= —= .053. 

100 

16i % = = .161 = .163+ 

*100 3 

Rule.—To express a rate per cent as a decimal, divide the 
number of per cent by 100. 

EXERCISE 96 

Express in decimal form: 

1 . 6%. 3 . 12%. 5 . 20%. 7 . 16%. 9 . 92%. 

2 . 10%. 4 . 15%. 6 . 35%. 8 . 14%. 10 . 83% 

208 


PERCENTAGE 


209 


How many times a number is : 

11. 100 % of it ? 14. 175 % of it ? 

12. 150 % of it ? 15. 300 % of it ? 

13. 200 % of it ? 16. 250 % of it? 

Express in decimal form: 

17 310%. 19. 540%. 21. 826%. 23. 130%. 25. 140%. 

18. 260%. 20. 375%. 22. 350%. 24. 225%. 26. 1000%. 

27. Eour times a number is wbat per cent of it ? 

28. Six times a number is wbat per cent of it ? 

29. Three and one fourth times a number is what per 
cent of it ? 

30. One and four fifths times a number is what per cent 
of it ? 

Express in decimal form : 

31.21%. 34. 5J%. 37. -1%. 40. .06%. 43.31%. 

32.41%. 35.871%. 38. .07 %. 41. .5%. 44.51%. 

33.31%. 36. 5|%. 39. 81%. 42. 5| %. 45.41%. 

Express in decimal form and observe the difference 
between: 

46. 5 % and .05 %. 49. 25 % and .25 %. 

47. 75 % and .75 %. 50. 11 % and .015 %. 

48. 2 % and .2 %. 51. i and i %. 

149. Expressing a Common Fraction as a Rate Per Cent. 

Example 1. Express ^ as a r^te per cent. 

Solution. 1. yt is T 5 T of or tt °i 100 °fo. 

2. .*. A = tt x 10° °t° ~ ¥r a °1° - 45 tt °1°- 

Example 2. Express as a rate per cent. 

Solution. 1. tt 5 3 = t¥t of or 100 °/o. 

2. iVV = t& x 100 °t° = 7 rsV °1° = 55 - 5 '#• 


210 


NEW HIGH SCHOOL ARITHMETIC 


Rule. — To express a common fraction as a rate per cent: 
Multiply the numerator by 100, and divide the result by the 
denominator. 

Note. — If the division is not exact, the quotient may be ex¬ 
pressed either as a mixed number, or be carried out to one or 
more decimal places as in Example 2 above. 

EXERCISE 97 


1 . Determine the rate per cent equal to the fractions in 
the following table; enter them in the adjoining column in 
pencil, and then memorize the results. 


1 


* 


* 


1*0 


i 


1 


* 


tV 


1 




1 


A 




* 


t 


TO 


1 


& 


i 


T2 



Using the results of the preceding table, give the rate 


per cent equal to : 


2. a. f; 

b. 2 


C * 2) 

d. 1J; 

e. U- 



3. a. j 

b. 

6 . 

c * TO-J 

d. y 8 ^; 

e - rf- 



4. a. JgL; 

b. 

15 

13. 

c - to> 

d. -J; 

«• Vw 



Determine the rate per cent equal to: 



5. f 

9. 

A 

13. 

32 

55* 

17 13 

- L/ - T2- 

21. 

16 

T3* 

6. J T . 

10. 

A- 

14. 

rf* 

18 - ft- 

22. 

9 5 
TTT* 

7 * A* 

11. 

xf* 

15. 

«• 


23. 

3 2 2 
45 0* 

8 - 

12. 

A* 

16. 

2 9 

T2* 

2 °- m ■ 

24. 

«• 

150. Finding the Percentage when the Base and 

the Rate are 


Given. 

Example 1. Eind 15 % of $ 38. 

Solution. 1. 15 % of $38 means .15 of $38. 

2. 15 of $38 = .15 x $38 = $5.70. 
















PERCENTAGE 


211 


Example 2. Find 121 % of $ 72. 

Solution . 1. 12^ % of $72 is the same as 1 of $72. 

2. .-. 12J °Jo of $72 = | of $72 = $9. 

Rule. — To find the percentage when the base and rate are 
given, multiply the base by the rate expressed in decimal form, 
or in fractional form. 


EXERCISE 98 

1. Find 1 % of 60, of 75, of 126, of 8409, and of 27,536. 

2. Find 2 % of each of the numbers in Example 1. 

3. Find 10 % of each of the numbers in Example 1. 

4. Find 25 % of each of the numbers in Example 1. 

5. Find 7 % of each of the numbers in Example 1. 

6. Find 15 % of $ 28, of $ 92, of $ 672, of $ 2344, and of 
$5476. 

7. Find 121 % of each of the amounts in Example 6. 
Also find 37i%, 621%, and 87^-%. 

8. Find 125% of $80. Also find 210%, 120%, and 
62^%. 

9. Find 31 % of $ 2500. Also find 4£ %, and 4J- %. 

10. Find 12i % of $ 1200. Also find 16| % and 200 %. 

11. A merchant sold goods that cost him $275 for 18 % 
more than they cost. What did he receive ? 

12. Out of 25 bu. of seed corn, only 75% germinated 
properly. How many bushels germinated properly ? 

How much is: 

13. 70 increased by 10 % of itself? By 30 % ? 

14. $ 2500 decreased by 20 % of itself ? By 50 % ? 

15. 35,000 sq. ft. of floor space increased by 121% 0 f 
itself ? 


212 


NEW HIGH SCHOOL ARITHMETIC 


16. $ 2650 increased by 12J % of itself ? Decreased by 
10 % of itself ? 

17. The employees in a certain factory were told that 
their wages were to be increased by 10 %. What would be 
the new wages of men who had been receiving: a. $ 85 ? 
b. $125? c. $140? d. $95? e. $175? 

18. 25 % of the 1500 trees planted on a real estate sub¬ 
division died. How many were left ? 

19. A man asking $ 8250 for his house and lot agreed to 
sell it for 5 % less to any one who bought direct from him 
instead of through an agent. What would be the difference 
in cost to the purchaser ? 

20. A grocer agreed to reduce all prices in his store 5 % 
for cash purchases and 3 % additional if the purchaser 
carried away his purchases instead of having them delivered 
by the grocer. What would be the new prices on goods 
which were marked: a. 75/? b. 40/? c. 82/? d. $1.75? 
e. 25/? 

21. The depositors in a certain bank that failed were 
promised 35 % of the amount of their deposits. How much 
should A, B, and C receive whose deposits were $785, 
$234.50, and $548.76 respectively? 

22. One plan of payments for bonds of the Fourth 
Liberty Loan provided that 10 % should be paid when the 
subscription was made, 20% on or before Nov. 21, 1918; 
20%, Dec. 19, 1918; 20%, Jan. 16, 1919; 30%, Jan. 30, 
1919. 

What were the payments for a person who subscribed for 
a $50 bond? For $3500 worth of bonds? For $400 
worth of bonds? 

23. A bankrupt paid his creditors 63 % of what he owed 
them. Wbat was received by a man to whom $ 1823.70 was 
due? 


PERCENTAGE 


213 


24. The salary of an agent was $ 1000 per year. He also 
received 3 % of the amount of all sales over $ 25,000. What 
was his total income during a year when he sold $ 73,000 
worth of goods? 

25. If 52 % of a certain ore is lead and If % of the 
balance is silver, how much of each metal is there in 10 tons 
of the ore ? 

26. A certain fertilizer is guaranteed to contain about 
4 % of nitrogen, 10 % of phosphoric acid, and 8 % of potash. 
How many pounds of each element are there in a ton of the 
fertilizer ? 


27. In a certain cow testing association, the three best 
producing cows and their records were: 


Cow 

No. 

Total Pounds 
op Milk 

Per Cent op 
Butter Fat 

No. Pounds 
op Fat 

Value of Fat 
at 30 < f > per Lb. 

1 

8109 

4.33 % 



2 

5023 

5.2 % 



3 

4897 

5.13 </o 




Complete the table. 

28. In brass, there is approximately 65 % of copper and 
35 % of zinc. How many pounds of each are there in a ton 
of brass ? 

29. How many pounds of nickel are required for 10 tons 
of nickel steel containing 2.8 % of nickel ? 

30. The employees in a certain company were informed at 
Christmas that their wages were to be increased by 15%. 
What would be the new daily wages of a man receiving: 
a. $1.75 per day? b. $3.25 per day? c. $6.40 per day? 

31. A contractor figures 2850 board feet of studding are 
required for a certain job. He decides to add 20 % for 
waste in cutting and for extras. How much should he 
order ? 












214 


NEW HIGH SCHOOL ARITHMETIC 


32. According to the rules of dealers, maple flooring of a 
certain grade may have not more than 7 % of the quantity 
ordered in lengths of 2 ft. to 3^ ft. How many feet of these 
short lengths may there be in an order for 4350 ft. ? 

33. A heating contractor observed that current low 
prices cause a decrease of about 15 % in the cost of install¬ 
ing a heating plant over the cost of two years ago. What 
rough estimate can he give of the cost of a plant similar to 
one that cost % 345.75 two years ago ? 

34. Architects charge a fee of 5 % for drawing the plans 
and writing the specifications for an ordinary dwelling. 
What will be the fee for a house costing $ 7568.45 ? 

35. 5 % of the energy supplied by a pulley belt to a line 
shaft by a 3-horse-power engine in a certain plant is lost be¬ 
cause of slipping of the belt. How much of the energy 
produced by the engine is actually delivered to the shaft ? 

36. Babbitt metal, used in making bearings for machinery, 
consists of about 80 % of tin, 10 % of copper, and 10 % of 
antimony. How much of each of these materials is re¬ 
quired to make 1500 lb. of Babbitt metal ? 

37. In the spring of 1932, farmers were asked to decrease 
by 10 % the number of acres planted to spring wheat in 
1931. How many acres should a man have planted who had 
planted 55 acres in 1931 ? 

38. Cloth loses 1-J % of its length when it is sponged. 
What will be the new length of a piece of cloth that 
measures 12 yards before it is sponged ? 

39. A foundryman making 47 lb. of bronze wants it to 
contain 40 % of copper, 10 % of tin, and the balance zinc. 
How many pounds of each metal are required ? 

40. In a certain company 1500 workmen are employed at 
an average wage of $ 2.75 per day. They ask for an in¬ 
crease of 15 % in their wages. How much would this 
increase add to the pay roll of the company ? 


PERCENTAGE 


215 


151. Finding what Per Cent one Number is of Another is the 

most common method of comparing two numbers. 

Example 1. 610 is what per cent of 2745 ? 

Solution. 1. 610 is the percentage, 2745 is the base, and the 
rate is unknown. 

2. Since the rate times the base equals the percentage, then 
the percentage divided by the base must give the rate. 

3. .*. the rate = 610 -f- 2745 = f = 22f %. 


Rule. — To find the rate when the percentage and the base 
are given, divide the percentage by the base. 


Note. — Usually it is wise to form a fraction whose numerator 
is the percentage and whose denominator is the base. If the 
fraction is easily reduced, it should be reduced, because the result 
may be one of the fractions given in Example 1, p. 210. If the 
fraction is not easily reduced, divide, carrying the result out, if 
necessary, to one or two decimal places, as in § 149. 


Example 2. 
Solution. 
Example 3. 
Solution. 


What per cent of 120 is 15 ? 
The rate = ^ = | = 12 \°Jo. 
37 is what per cent of 150 ? 
The rate = = 24.6 °/o. 


EXERCISE 99 

1. What per cent of 24 is 12? 6? 8? 4? 2? 3? 

2. What per cent of 30 is 3 ? 6 ? 5 ? 15 ? 10 ? 

3. What per cent of 48 is 16 ? 6 ? 24 ? 36 ? 

4. What per cent of 50 is 5 ? 20 ? 30 ? 25 ? 10 ? 

5. What per cent of 72 is 9? 18? 48? 36? 144? 

6. What per cent of 36 is 24 ? 48 ? 72 ? 90 ? 360 ? 

7. 3 is what per cent of 12 ? of 9 ? of 24 ? of 30 ? 

8. 11 is what per cent of 33 ? of 66 ? of 44 ? of 55 ? 

9. 14 is what per cent of 35 ? of 42 ? of 28 ? of 84 ? 

10. 32 is what per cent of 48 ? of 64 ? of 320 ? of 160 ‘l 

of 16? 


216 


NEW HIGH SCHOOL ARITHMETIC 


11. What per cent of 25 is 5 ? is 25 ? is 50 ? is 100 ? 

12. 40 is what per cent of 60 ? of 20 ? of 50 ? of 90 ? 

13. 22 is what per cent of 11 ? of 44 ? of 33 ? 

14. What per cent of 64 is 16? is 64? is 128? is 96? 

15. 35 is what per cent of 25 ? of 350 ? of 20 ? of 5 ? 

16. What per cent of 200 is 50? is 140? is 250? is 175 ? 

17. An article is composed of 19 parts of silver and 6 
parts of copper. What per cent of the whole is the silver ? 
What per cent is the copper? 

18. From a barrel of gasoline containing 50 gal., 12 gal. 
were lost by leakage. What part was lost ? 

19. A certain oil company found that its pump for 
retailing gasoline to motorists was delivering -J- pt. too 
much on every 5-gallon sale. a. The excess delivered was 
what per cent of the amount sold ? b. At that rate, what 
was the loss in gallons in a day when 2468 gallons were 
sold ? 

20. Out of a population of 26,832 in a certain town, 3354 
were foreign born. What per cent were foreign born ? 

21. If 675 lb. of silver are obtained from 45 T. of a 
certain ore, what per cent of that ore is silver ? 

22. A boy in a machine shop spoiled 8 bolts out of every 
200 that he made. What per cent did he spoil ? 

23. A certain shop that employed 275 men and boys in 
1929 had work for only 25 employees during the year 1932. 
What percentage of the number employed in 1929 was the 
number employed in 1932 ? 

24. Laborers who were earning $4.50 per day in 1929 
were able to get only $ 3 per day in 1932. 

a. What per cent of their 1929 wages was their 1932 
wages ? 

b. What per cent of their wages in 1929 was the decrease ? 


PERCENTAGE 217 

25. Round steak sold for 55 $ per pound in 1929 and for 
25 ^ per pound during the summer of 1932. 

a. What per cent of the selling price in 1929 was the de¬ 
crease between then and 1932 ? 

b. What per cent of the 1929 selling price was the 1932 
selling price ? 

26. Anthracite coal sold for $ 9 per ton in 1916 and for 
$ 16 in 1932. What per cent of the 1916 price was the 
1932 price? 

27. Milk sold for 12 per quart in Chicago in 1918 and 
for 9 1 per quart in 1932. What per cent of the 1918 price 
was the decrease? 

28. A contractor finds that 225 board feet of flooring out 
of an order of 2500 board feet is in pieces less than 3i ft. 
long. Specifications were that not over 7 % were to be in 
short lengths. Was the flooring up to specifications ? 

29. A farmer bought some seed corn, the germination of 
which was said to be 93 %. He planted three separate lots 
of 50 seeds each. In pan A, 42 seeds germinated j in pan 
B, 40 seeds germinated; in pan C, 44 seeds germinated. 
What was the percentage of germination in each pan? 
What was the average percentage of germination? Was 
the seed up to its advertised grade ? 

30. The following table gives the average pay per hour 
of certain union workmen in 1916 and 1930. Complete the 
table. 


Workmen 

Pay per 
Hour in 
1916 

Pay per 
Hour in 
1930 

Increase 

Per Cent 
op 

Increase 

Bricklayer . . . 

65 i 

$1.69 



Carpenter . . . 

57jzf 

1.39 



Cement Finisher . 

62^ 

1.49 



Painter .... 

62^ 

1.46 













218 


NEW HIGH SCHOOL ARITHMETIC 


31. Out of 240 eggs placed in an incubator, 20 were found 
infertile. What per cent were infertile ? 

32. Out of 240 eggs placed in an incubator, 190 chicks 
were hatched. What per cent of the eggs “ hatched out ” ? 

33. One mixture of concentrated feed for cows is: 

Wheat bran, 100 lb. Gluten meal, 75 lb. 

Ground oats, 100 lb. Dried brewers’ grain, 75 lb. 

Corn meal, 150 lb. Cottonseed meal, 50 lb. 

What per cent is each constituent of the total ? 

34. A molder put up 168 molds in a day, each containing 
16 pieces. When he sorted the completed work, 2544 
pieces were found to be perfect. What per cent of his 
work was perfect ? 

35. The weight of corn on the cob is 72 pounds, and of 
shelled corn is 56 pounds. What per cent of the weight 
on the cob is the weight of the cobs? 

36. Steel is made by heating iron and other metals in a 
furnace until they melt and fuse. 9 tons of iron and ^ ton 
of the other metals were placed in a furnace. 8^ tons of 
steel were obtained. What per cent of the material was 
lost in the process ? 

37. If a man who was earning $ 135 per month had his 
pay raised to $150 per month, what was the per cent of 
increase ? 

38. The entire length of the Panama Canal is 50.5 miles. 
Its length on land is 40.5 miles. What per cent of its 
length is on land ? 

39. A contractor took a job for a contract price of $2644. 
He found that the work and material cost him $2350. 
What per cent of the sums he spent were his profits ? 

40. The population of Birmingham, Alabama, in 1920 
was 178,806 and in 1930 was 259,678. What was the per 
cent of increase ? 


PERCENTAGE 


219 


41. During an epidemic of “ colds,” 50 pupils out of an 
enrollment of 450 in a school were absent one day. What 
per cent of the pupils of the school were absent on that day ? 

42. Mr. Edward’s salary was decreased from $360 per 
month to $ 310 per month. What per cent of his salary 
was the reduction? 

43. John had been receiving 50^ per hour. His em¬ 
ployer had to cut him to 40 ^ per hour. 

a. What per cent of his former pay per hour was his new 
pay? 

b. What per cent of his former pay was the “ cut ” ? 

44. During the winter of 1932 there was a general de¬ 
mand for reduction in taxes. In one county of a certain 
state, the taxes collected in 1932 amounted to $ 375,500. 
The taxes levied for 1933 amounted to $ 315,000. 

a. What reduction in taxes was made in this county ? 

b. What per cent of the taxes levied for 1932 were the 
taxes levied for 1933 ? 

45. A merchant sold $18,500 worth of goods during 
December, 1931, and $ 16,250 worth during December, 1932. 
What per cent of his sales in December, 1931, was the 
decrease ? 

46. On an invoice amounting to $ 285, a purchaser was 
given a reduction of $5.70 for prompt cash payment. 
What per cent of the bill was the reduction? 

47. The population of the United States in 1920 was 
105,710,620 and in 1930 was 122,775,046. What per cent 
of the population in 1920 was the increase? 

48. The population of the United States in 1910 was 
91,972,226. 

a. What per cent of this number was the increase from 
1910 to 1920 ? 

b. How did the per cent of increase from 1910 to 1920 
compare with the per cent of increase for 1920 to 1930 ? 


220 


NEW HIGH SCHOOL ARITHMETIC 


49. The number of paying passengers carried on rail¬ 
roads in each of certain years is given in the following table. 
Complete the table. 


Year 

Paying Passengers 

Decrease 
from 1920 

Per Cent Decrease 
from 1930 

1920 

1,270,000,000 



1925 

902,000,000 



1928 

798,000,000 



1930 

710,000,000 




Suggestion. Round off all the numbers by dropping the last 
six zeros. 

50. The three states of the United States in which the 
rate of increase in population between 1920 and 1930 was 
largest were California, Florida, and Michigan. 


State 

Population 

1920 

Population 

1930 

Increase 

Rate of 
Increase 

California . . 

3,426,861 

5,677,251 



Florida . . . 

968,470 

1,468,211 



Michigan . . 

3,668,412 

4,842,235 




a. Find the increases over the 10-year period. 

b. What per cent of the population in 1920 in each state 
is the increase? 

51. The public debt of the United States was largest in 
1920, when it reached $24,330,889,731. By 1931, this 
debt had been reduced to $ 16,481,024,525. 

a. Round off each of these figures to millions of dollars. 

b. Find the decrease in millions of dollars. 

c. What per cent of the debt in 1920 (as rounded off in 
part a) was the decrease ? 


















PERCENTAGE 


221 


152. Finding the Base when the Percentage and the Rate are 
Given, while less common a problem than those considered 
in § 150 and § 151, is necessary at times. 

Example 1. 85 is 5 % of what number ? 

Solution. 1. 85 is the percentage, 5 °Jo is the rate per cent, and 
the base is unknown. 

2. Since the rate x the base = the percentage, 
then the base = the percentage -f- by the rate. 

3. .\ the base = — = 1700. 

.05 

Check. Does .05 x 1700 = 85? Yes. 

Rule. — To find the base when the percentage and the rate 
are given, divide the percentage by the rate. 

Note. — If the rate is easily expressed as a common fraction, 
the solution may be performed as in Example 3, below. 

Example 2. What number increased by 8 % of itself is 
540? 

Solution. 1. The number is 100 °Jo of itself. When increased 
by 8 <jo of itself, the result is 108 °/o of the number. 

2. .-. 540 = 108 °Jo of some number. 540 is the percentage and 

108 °]o is the rate per cent. 

3. .\ the base = 540 -4-1.08 = 500. 

Check. Does 1.08 x 500 = 540? Yes. 

Example 3. 126 is 87-1 % of what number ? 

Solution. 1. 87£ °/o = ■£. 

2. .*. 126 = | of some number. 

3. | of the number = } X 126 = 18, 
and f of the number = 8 x 18 = 144. 

EXERCISE 100 
Find the number of which : 

1. 20 is: a. 2%; 6. 5%; c. 8%; d. 40%; e. 50%. 

2. $56 is : a. 7% ; 6. 6% ; c. 10 % i d. 28 % ; e. 42 %. 


222 


NEW HIGH SCHOOL ARITHMETIC 


3. 6 % of what number is : 

a. 39 mi.? b. $45? c. 231 T.? 

4. 15 J 0 of what number is : 

a. 630? b. $91.50? c. $439.50? d. 750? 

5. Of what number is : 

a. 75 rd. 25 J 0 ? b. $240 6 % ? c. 16 T. 66} Jo ? 

6. What number diminished by 36 Jo of itself is 420 ? 

7. If the expenses of collection of a charity fund are 
about 8 %, approximately how much should be collected in 
order that $ 1500 may be left ? 

8. It is said that only 60 J 0 of an army actually fight at 
the front, the remaining 40 J 0 being needed to take care of 
these 60 J 0 . How large must an army be therefore in 
order to have 1,500,000 men available for actual fighting ? 

9. A man said that his taxes for a certain year were 
$ 131.25. If this is If J 0 of the assessed valuation of his 
property, what was the valuation placed upon his property ? 

10. A man receives from a source other than his salary 
an income of $ 2100. Of what sum is this 6 J 0 ? 

11. A poultryman is getting a good “ hatch ” if 85 J 0 of 
the eggs that he places in an incubator hatch out chicks. 
How many eggs must be placed in incubators in order to 
obtain about 500 chicks ? 

12. Since cloth loses about 1| J 0 of its length when it is 
sponged, how much cloth must a woman buy in order to 
have 18 yards full length after sponging ? 

13. The population of Tacoma, Washington, in 1910, was. 
83,743. This was said to be 122 J 0 of its population in 
1900. What was its population in 1900 ? 

14. The population of Oklahoma City, Oklahoma, in 
1910 was 64,205. This was said to be 539.7 J 0 of its 
population in 1900. What was its population in 1900 ? 


PERCENTAGE 


223 


15. There is a shrinkage of about 15 % in the ice placed 
in an ice house, before it is sold during the following 
summer. How much ice should a company store in order 
to have at least 25,000 tons to sell during the summer 
season ? 


EXERCISE 101 
Miscellaneous Exercises 

1. Eind the missing numbers in the following table: 


p 


B 

435 

15% 

? 

508.25 

? 

2675 

? 

6f °fo 

5278 

76 

? 

294 

322 

4 \°fo 

? 

? 

3f °/o 

15,275 


2. A merchant has floor space of 40 ft. by 70 ft. He 
decides that he must have about 35 % more floor space. 
How much floor space will he then have all together ? 

3. A man paid $ 273.45 in taxes last year. He learns 
that taxes are to be increased about 8 °fo • What is the 
probable amount of his taxes for the coming year ? 

4. In the month of May, 18 days were cloudy. What 
per cent of the days were cloudy ? 

5. In one class, 14 pupils out of 27 were marked 
excellent. In another class, 4 pupils out of 23 were marked 
excellent. Eind the per cent marked excellent in each 
class. 

6. A man whose salary was $2750 per year was given 
an increase of $ 250. What per cent of increase was this ? 









224 


NEW HIGH SCHOOL ARITHMETIC 


7. A contractor’s bid on a certain street construction job 
was $ 55,785. It was agreed that 10 % of the amount was 
to be retained by the town for two years as a guarantee 
fund, to insure compliance with the specifications. How 
much would the contractor receive upon completion of the 
job? 

In Examples 8 to 20, do the computation mentally. Give 
the results orally or write only the results. 

8. What is 10% of: a. 95? b. $2750? c. 3625 bu. ? 

9. What is 12i % of: a. 64? b. $320? c. 480 mi. ? 

10. What per cent of 64 is : a. 16 ? b. 32? c. 4 ? 

11. $ 16 is 20 % of what number ? 

12. What is 16-J % of : a. $75? b. 360 sq. rd. ? 

13. What per cent of $ 48 is : a. $ 12 ? b. $ 16 ? 

14. Of what number is $ 2.75 five per cent? 

15. If the enrollment in a school changed from 200 to 
220, what per cent of the first enrollment was the increase ? 

16. An overcoat marked $27.50 was sold for $’25. What 
per cent of the former price was the decrease ? 

17. What must be paid for a suit if it is marked $ 15 
less 25 % ? 

18. If the pay of a workman is cut from 45 $ per hour to 
35 $ per hour, what is the rate per cent of decrease ? 

19. One winter, eggs sold for 24 $ per dozen. A year 
later they sold for 36 / per dozen. What per cent of the 
former price was the increase ? 

20. A house and lot bought for $ 8000 was sold for 
$ 7500. What per cent of the cost was lost? 

21. When gas is made from coal in a gas factory, coke is 
obtained as a by-product. The amount of coke obtained is 
about 70 % of the amount of coal used. How many tons of 
coke will be produced in a factory which consumes 3750 
tons of coal per month ? 


PERCENTAGE 


225 


22. A contractor figures that the actual cost of a certain 
job is $13,275. He decides to add 10 % for his profits and 
5 °jc for possible but unforeseen extra expenses. How much 
will his bid be ? 

23. If I buy a piece of property for $ 6500, what must I 
sell it for in order that I may receive 15 % more for it than 
I pay? 

24. A merchant paid $ 17.50 for a table. He marked it 
for sale at $ 22.50. What per cent of the cost price had he 
added to his cost price to secure his selling price ? 

25. The enrollment in the senior class of a school is 125. 
There had been 275 pupils in the class when they entered 
the first year of the high school. 

a. What per cent of the first-year class are in the senior 
class ? 

b. What per cent of the first-year class have dropped out 
during the first, second, and third years ? 

26. The state of New York gained 2,202,839 in popula¬ 
tion between 1920 and 1930. This was an increase of about 
21 % of the population in 1920. What was the population 
in 1920 ? 

27. Texas gained 1,161,487 between 1920 and 1930. 
This was about 25 % of the population in 1920. What was 
the population in 1920 and in 1930 ? 

28. The population of the South Atlantic States increased 
1,803,317 between 1920 and 1930. This was 12.9 % of the 
population in 1920. Rind the population in 1920 and also 
in 1930. 


VIII. BUYING AND SELLING GOODS 


153. The advertised price at which an article may he 
bought is called the List Price. Sometimes the article is 
sold for less than the list price ; then the difference between 
the list price and the actual selling price is called a Dis¬ 
count. 

A discount is usually stated as a certain per cent of the 
list price. 

Thus, a discount of 2 °Jo from the list price means that 2 <fc of 
the list price is subtracted from the list price. 

154. Why Discounts are Given. — Wholesalers usually pub¬ 
lish a general catalogue describing their goods. In this 
catalogue, a list price for each article is given. This price 
is usually made great enough so that it is the maximum 
price that must be charged even if the price of material and 
labor should increase. In order to inform their customers 
about the actual prices at any time, wholesalers issue from 
time to time a list of discounts from these published list 
prices. In rare cases, especially recently, wholesalers have 
had to add to their list prices in order to arrive at adequate 
selling prices. 

The following is an excerpt from such a price list, giving 
prices relating to page 184 of a certain jobber’s catalogue. 

Page 184. Hand Power Punches and Shears 


Fig. A 1 . . List plus 60 °Jo. Fig. A 10 . . List plus 60 % 

Fig. A3. . List plus 60 °Jo. Fig. A 42 . . List plus 60 % 

Boilermakers’ Screw Punch.Discount of 45 °fo 

Extra parts for same.Discount of 50 % 

Fig. 510 B, Simplex Shear.Net 

Beading Tool, each . ..$.90 


226 






BUYING AND SELLING GOODS 


227 


Sometimes special discounts are given to favored cus¬ 
tomers, either because they are large purchasers or as a 
special courtesy to them because of personal friendship. 

155. Sometimes More Than One Discount is offered. In 
such cases, the first discount is computed on and subtracted 
from the list price ; the second discount is computed on and 
subtracted from the first remainder; etc. 

Example . Find the net cost of an automobile tire listed 

at $ 32.60, with discounts of 20 % and 10 °/ 0 - 

Solution. 1. The list price of the tire = $ 32.60 

2. 20 % or } of $32.60 = 6.52 

The first remainder = $ 26.08 

3. 10 <fo or X V of $ 26.08 = 2.61 

The net price = $ 23.47 

156. When goods are bought, an agreement is made 
about the time when payment is due. This comes under 
the Terms of Sale. The sale may be for “ net cash/ 7 or for 
“ net cash in 10 days/ 7 or 30 days, or 90 days. When pay¬ 
ment is not demanded at once, a discount for cash payment 
is usually offered. 

Thus, the terms of a sale may be “2 °fo for cash in 10 days, or 
net cash in 30 days.” The cash discount is subtracted from the 
net cost after all other discounts have been subtracted. 

157. Below is a sample Invoice, modeled after one used 
by a well-known manufacturing company. Note in it the 
manner of indicating the terms of payment, as well as the 
general form of the invoice. 

In this invoice, “2%/l^th Prox.” means that 2 °fo will be 
allowed if payment is made on or before the 15th of the next 
month. 

At the end of the invoice, the 2 °Jo cash discount had been de¬ 
ducted by the clerk at Wharton and Huber’s at the time of mak¬ 
ing payment, for, as indicated by the Receipt, the payment was 
made before the 15th of the next month* 



228 


NEW HIGH SCHOOL ARITHMETIC 


The Stowe Manufacturing Co. 

, c 'fre/vu. 6, 19c 32 

Brighton, Mass., _/_ _, 


Sold to _ 
Address 


7AMia,\Z(yyi V ft-'uk&v, 

2^3 jokwfro-w o ft., 

?V{aAf'witt&, <Jlt. 


Terms:. 


2°/ 0 I/6tk gvooc. 


_Cases_lbs. 

_Pkgs-lbs. 


Shipped by_ 

2j 8 w-Ltlv otA&v 


CfOO-cUs. 


Quantity 

List No. 

Description 

Price 

Extension 

Total 

2i dz. 

J 364 

Desk Locks 

Less 16f, 7 \°fo 

$6.38 dz. 

15 

95 

12 

29 

l|dz. 

O 426 

Bronze padlock 
Less 10 °Jo 

21.70 dz. 

27 

13 

24 

42 



Parcel post 





17 



Insurance 





05 







36 

93 



2 °Jo c 

ash discon 

nt 



73 



Received Pa 

yment 



36 

20 



Mar. 8, 1 

932 







Stowe 

Man’fg. C 

0. 





Note. — The symbol N/30 or N/60 means “ net cash in 30 days ’ 5 
or in 60 days respectively. 


EXERCISE 102 

Find the net price when the rate of discount is: 
a. 10 % b. 20 % ; c. 50 % ; d. 25 % ; 

e. 331 % ; f. 5%; g. 12|% ; h. 16| % ; 

and the list price is : 

1. $15.00. 3. $2.35. 5. $336.00. 7. $47.60. 9. $4.15. 

2. $25.00. 4. $7.86. 6. $125.00. 8. $27.50. 10. $8.72. 




























BUYING AND SELLING GOODS 


229 


Find the net amount of an invoice for: 

11. $ 60 less 331 % a nd 10 %. 

12 . $ 80 less 10 % and 331 %. 

13. $ 5.45 less 20 % and 10 %. 

14. $ 15.84 less 121 oj 0 an d 25 

15. $ 27.65 less 20 % and 5 %. 

16. $ 18.50 less 66f % and 5 %. 

17. $ 250 less 121 % and 10 %. 

18. $ 563.20 less 121 % and 15 %. 

19. $ 26.75 less 20 10 %, and 5 

20. $ 73.75 less 40 20 and 15 


Find the net amount of the following invoices : 


Ex. 

No. 

Cost at 
List Price 

Trade 

Discount 

Terms 

Date of 
Invoice 

Date Paid 

21. 

$54.90 

10 <f 0 

2/10; N/30 

June 8 

June 30 

22. 

231.00 

20%, 10% 

5/30; N/90 

May 10 

June 1 

23. 

19.60 

25 %,12 \ % 

1/10; N/30 

Sept. 12 

Sept. 20 

24. 

125.85 

331 %, 5% 

3/30; N/90 

Aug. 6 

Sept. 20 

25. 

5265.00 

331 25 % 

5/30; N/60 

Apr. 13 

May 10 


Note. — Other drill examples can be made by inserting other 
prices for those given in Examples 21-25. 


26. A local hardware dealer gives 5 % discount for cash. 
Find the amount to be paid on the following purchases for 
cash : 

1 garden spade @ $ 1.85 ; 1 garden hoe @ $ .60; 1 gar¬ 
den rake @ $ 1.10; 1 hand cultivator @ $ .20; 1 large 
cultivator @ $ 1.45. 













230 


NEW HIGH SCHOOL ARITHMETIC 


The following examples were copied from actual invoices 
received by a hardware dealer. Find the net amount of 
each invoice. 

27. Item A, amounting to $116.40, less 87^-%, 25%, 

25%, 5%. 

Item B, amounting to $ 79.00, less 70 %, 5 %, 20 %, 

10 %. 

Item C, amounting to $ 10.95, less 85 %, 10 %, 10 %, 
5%. 

28. Item A, amounting to $ 28.00, less 85 %, 25 %, 25 %, 

5%. 

Item B, amounting to $202.30, less 87^-%, 25%, 
25%, 5%. 

Paid so as to receive also a cash discount of 2 %. 

29. Item A, amounting to $ 4.50, less 85 %, 10 %, 10 %. 
Item B, amounting to $ 22.50, less 85 %, 25 %, 25 %, 

5%. 

Item C, amounting to $209.20, less 87|- %, 25%, 
25 %, 10 %. 

30. Item A, amounting to $ 42.50, less 62-1- % ? 5 %, 25 %, 

10 %. 

Item B, amounting to $ 7.50, less 85 %, 10 %, 10 %, 
5%. 

31. Determine the amount to be paid by a firm for two 
#82 keys listed at 25^ each, less 30% and 15 %, assum¬ 
ing that they paid the bill in time to secure an additional 
discount of 2 % for cash. 

32. The following approximate prices are quoted on ordi¬ 
nary garden hose by one dealer : 

A grade, in 500-ft. rolls, 35 % and 5 % off list price. 
Less than 500 ft., 30 % off list price. 


dkM 


* 


BUYING AND SELLING GOODS 


231 


B grade, 25 % off list price. 

C grade, 50 °] 0 off list price. 

Determine the approximate cost of: 

1500 ft. f " A grade hose listed at 25 ^ per foot. 

1200 ft. A grade hose listed at 19 ^ per foot. 

1000 ft. f 1 " B grade hose listed at 19 $ per foot. 

2500 ft. !" C grade hose listed at 15 $ per foot. 

33. Determine the cost of the following blacksmith’s 
supplies bought for cash. 

One ft 10 heavy duty anvil listed at $ 110.00 hut sold at 
an advance of 15 % over list price. 

One ft 104 5-lb. sledge listed at $ .30 per pound, less 50 %. 

One ft 103 3-lb. double-faced hammer, listed at $ .40 per 
pound, less 45 %. 

One round blacksmith’s punch, -J", weighing 1 lb. 12 oz. 
at $ .25 per pound. 

One round blacksmith’s punch, |", weighing 2 lb. 4 oz., 
at $ .25 per pound. 

One 10" blacksmith’s pincers, listed at $ .75, less 10 %. 

One 12" blacksmith’s pincers, listed at $ .86, less 10 %. 

One 14" blacksmith’s pincers, listed at $ 1.00, less 10 %. 

Terms : 5 % cash, or N/30 days. 

Make out in full an invoice from each of the following 
sets of data, patterning after the sample on page 228. Use 
this year in dates. 

34. The United States Book Co., Chicago, Ill., sold to 
Hendricks and Son, Danville, Mich., on Aug. 14: 45 copies 
Beginners’ Algebra @ $ 1; 45 copies First Course in English 
@ $ 1.25; 12 copies High School Physics @ $ 1.25. Allow 
a dealer’s discount of 20 %. Terms, 2 % in 10 days, net in 
30 days. Assume that Hendricks and Son pay the bill on 
Aug. 21. How much money should they send ? 


232 


NEW HIGH SCHOOL ARITHMETIC 


35. The Union Tire Co. of Akron, Ohio, on July 23, 
sold to J. B. Friend and Co., of Summit, Ohio: 12 Clincher 
casings 30 x 31 @ $ 20.45; 10 Clincher casings 30 x 31 @ 
$ 17.67; 12 Dunlop casings 32 x 31 @ $ 24.38; 6 Dunlop 
casings 33 X 4 @ $ 31.25; 6 Dunlop casings 34 x 4 @ 
$ 36.85. 

Prices subject to discounts of 12^-% and 10%. Terms, 
5 % off for cash before 10th of next month. Bill paid 
Aug. 5. Determine amount paid. 

36. The Natenas Enameling Co., Chicago, Ill., on Mar. 16, 
sold to David Jenkins and Co., of Hastings, Indiana, 
4i doz. # 200 stove pans @ $ 6 per dozen, less 40, 10, 5, 
and 5 %. They made a cash discount of 25 on the whole 
bill to cover the freight. Terms, 2 % for cash in 10 days, 
net in 60 days. The bill was paid on Mar. 27. Find the 
amount. 

37. The Acme Screw Co., Providence, R. I., sold to the 
Kerwin Hardware Co. of Montrose, N. Y., on May 12, 
the following: 10 gr. each of the following bright flat-head 
screws, f" #8 ® $ .95; 1" ft 6 @ $ .80; l\" #10 @ $ 1.40 ; 
11" #11 @ $1.50; 11" #12 @ $1.65; 11" #10 @ $1.30; 
li" #12 @ $1.55. These were sold at list less 87-1, 5 ? 25, 
and 10 %. (These were the actual discounts.) Also 10 gr. 
each of the following blued round-head screws : 21" #14 @ 
$2.90; li" #12 @ $1.65. These were sold at list less 85, 
5, 25, and 10 %. Also 10 gr. nickel round-head screws 
li" #8 @ $1.30 less 771, 5 , 25, and 10 %. 

The terms were 2 % for cash in 10 days, net in 30 days. 
The bill was paid on May 25. Find the amount. 

38. The Manville Manufacturing Co., Chicago, Ill., sold 
to A. B. Stephens of Kewauna, Ohio, on Aug. 14, one 
5 H.P. stationary gasoline engine, listed at $ 178, with 
$ 10 extra for a kerosene attachment. It was sold at list 


BUYING AND SELLING GOODS 233 

less 30 %. Terms, 3 % for cash in 10 days, net in 30 
days. The bill was paid on Aug. 29. Find the amount. 

39. The Winston Garage bought the following supplies 
from the Central Auto Supply Co. of Cleveland, Ohio, on 
Feb. 24 : \ doz. automobile jacks ft 6, listed at $ 2, less 25 % ; 
3 doz. ft 20, size 93 grease cups @ $ .40 each, less 75 % ; 
i doz. Ford Tool Sets at $ 3.73 each, less 20 % and 10 %. 
Terms, cash in advance. 

40. J. B. DeLancey, a street contractor, ordered from 
the Spencer, Baldwin Co. of Detroit, Mich., on June 5, the 
following tools: 2 doz. Fig. 57 C, #4, D handle shovels @ 
$ 19.50 per doz., less 10 %; 1 doz. Fig. 50 B, ft 3, D handle 
shovels @ $ 19, less 5 % ; 2 doz. Fig. 85, 8 lb. clay picks @ 
$ 16, less 35 % ; 3 doz. Fig. 95 pick handles @ $ 11.50, 
less 50 <J 0 ; 6 Fig. 52 wheelbarrows @ $ 66 per dozen, less 
5 % ; 3 scrapers, size 2, @ $ 8.25, plus 65 %. Terms, 5 % 
off for cash in 30 days, net in 90 days. Bill paid July 1. 
Find the amount. 

41. B. M. Chandler and Co. publish the following changes 
in prices from those published in their Catalogue ft 27. 

Polished Brass Bow Locks, Style 105, list plus 80%. 

Polished Brass Bow Locks, Style 103, list plus 40 %. 

Spruce Boat Oars, per foot 16 ft. 

Terms, 2 %/10 days, N/30 days. 

Determine the cost to the Lake View Boat Co. of the fol¬ 
lowing goods, bought on April 15 and paid for on April 22. 

6 pairs #2, Style #105, polished brass row locks, listed at 
$2; 12 pairs #2, Style #103, polished brass row locks, 
listed at $1.75; 6 pairs 7 ft. spruce oars; 6 pairs 1\ ft. 
spruce oars; 6 pairs 8 ft. spruce oars. 


234 


NEW HIGH SCHOOL ARITHMETIC 


Profits and Losses 

158. Profits and Losses in business transactions are often 
described as per cents of the cost or of the selling price. 

The Cost of an article is the actual sum paid for it. 

The Gross Cost of an article is the cost plus the other ex¬ 
penses incurred in securing and marketing the article. 

The Selling Price of an article is what the merchant re¬ 
ceives for it. 

If the selling price is more than the cost plus the ex¬ 
penses, then the merchant makes a Profit. Therefore, 

profit = selling price — cost — expenses 

Merchants call the difference between cost and selling 
price their Margin. 

Profits on Sales and Margins are generally expressed now 
as a per cent of the selling price. 

Thus, if a table cost $10 and is sold for $15, the margin is $5. 
Out of this margin, however, the merchant must subtract the cost 
of securing and handling the table before he can tell what his 
profit on the sale of the table really is. If such expenses amount 
to $3, then there remains out of his margin a profit of $2. This 
profit is | or 20 °Jo of the cost of the table; it is or 13£% of the 
selling price. 

To make a profit, a merchant must add to his cost enough 
to pay his expenses and also his profit. Some merchants do 
this by adding to the cost a certain per cent of the cost; 
others by adding to the cost a certain per cent of the selling 
price. 

Example 1. An article cost $ 8.00. For what must it be 
sold to have a margin of 20 % of the cost? 

Solution. 1. 20 °/o of $8.00 == $1.60. 

2. .-. the S. P. = $8.00 + $1.60 

= $9.60. 


INTEREST 


235 


Example 2. An article cost $8.00. What must the sell¬ 
ing price be in order that the margin shall be 20 % of the 
selling price 9 

Solution. 1. The S. P. is 100% of itself. The margin is 20% 
of the S. P. the cost must be 80% of the S. P. 

2. 80% of S. P. = $8.00. 

3. S. P. = $8.00 - .80, or $10. 

Check. 20% of $10 = $2, the margin. 

$10 - $2 = $8, the cost. 

EXERCISE 103 


Find the selling price if the margin is the per cent of the 
cost given in Examples 1 to 10. 


Ex. No. 

Cost 

Rate op 
Margin 

Ex. No. 

Cost 

Rate op 
Margin 

1 . 

$2200 

10% 

6. 

$625 

20% 

2. 

1650 

20% 

7. 

1040 

12*% 

3. 

750 

5% 

8. 

6400 

50% 

4. 

1460 

6% 

9. 

3225 

10% 

5. 

2100 

33i% 

10. 

4500 

25% 


Examples 11 to 20 . Use the costs and rates of margin 
given in Examples 1 to 10 to find the selling price, but con¬ 
sider the margin as a per cent of the selling price. 

In Examples 21 to 30, find the rate of gain or loss as a 
per cent of the selling price. 


Ex. No. 

Cost Was 

Selling 
Price Was 

Ex. No. 

Cost Was 

Selling 
Price Was 

21. 

$125.00 

$150.00 

26. 

$32.50 

$35.75 

22. 

64.00 

72.00 

27. 

450.00 

375.00 

23. 

150.00 

135.00 

28. 

8.00 

11.00 

24. 

275.00 

200.00 

29. 

1275.00 

1500.00 

25. 

1050.00 

1200.00 

30. 

2650.00 

2500.00 
























236 


NEW HIGH SCHOOL ARITHMETIC 


31. An article was bought for $ 8 and sold for $ 12. The 
selling expenses were $ 2. 

a. What was the margin ? b. What was the profit ? 

c. What per cent of the cost was the profit ? 

d. What per cent of the selling price was the margin ? 

32. Repeat Example 31 if the cost was $ 15, the selling 
price was $ 25, and the selling expenses were $ 4. 

33. A dealer purchased goods for $ 845 and sold them at 
a loss of 33^ %. What was his loss ? 

34. A tradesman sold at a loss of 25 % goods that cost 
$ 586.20. What did he receive ? 

35. A dealer bought 25 barrels of flour at $ 4.40 per 
barrel. At what price must he sell them per barrel to have 
a margin of 15 % of the cost ? 

36. A merchant sold some goods for $416.50. If it is 
assumed that he marks his goods so as to have a margin of 
20 °f 0 of the selling price, what did the goods cost ? 

37. A wholesale dealer bought 250 bu. of potatoes at 
$ .33 per bushel. The freight was $ 8.50, and the cartage 
was $ 5.25. At what price must he sell them per bushel so 
as to gain 15 °/ 0 on the gross cost ? 

38. What per cent of the cost is the profit if a lot is 
bought at $ 420 and sold for $ 595, if the seller has to pay 
an agent a commission of 5 % of the selling price ? 

39. What per cent of the selling price is the margin if 
peaches are bought at $ 3.25 per bushel and sold for $ .25 
per quart ? 

40. A man bought a lot costing $ 525 and built upon it a 
house costing $ 2750. He had to pay $ 235 for special im¬ 
provements on the street. He sold the house and lot for 
$ 3775. What per cent of this total cost was the difference 
between this cost and the selling price ? 


BUYING AND SELLING GOODS 


237 


Marking Goods 

159 . Merchants often employ a secret system for mark¬ 
ing the cost and selling price of their goods. Often the 
letters of some word are used to represent the digits with 
which numbers are written. 


Thus, the letters of the word manuscript , 
adjoining, might be used to represent the 
digits below them respectively. 


manuscript 

1234567890 


Then, to represent $ 269, the letters acp would be written on 
the price tag. 


To avoid the repetition of one letter where a price like 
44 ^ is to be represented, a repeater such as the letter z is 
used. 

Thus, 44 ^ would be represented by uz. 

A typical price tag representing a cost of 
$13.57 and a selling price of $16.75 is shown 
at the right. 


- 5 - 

mnsr 

mcrs 


160 . One problem of the merchant is the determination 
of a selling price so as to gain a reasonable profit on the 
cost. 

Example. A merchant wished to mark an article cost¬ 
ing $ 1.28 at an advance of 25%. Determine the selling 
price, and represent it and the cost by the letters of the 
word manuscript. 

Solution . 1. 25 °fo of $ 1.28 = $ .32. 

2. the selling price = $ 1.28 + $ .32 = $ 1.60. 

3. The cost will be represented by mai 

and the selling price by met. 




238 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 104 

Determine the selling price and represent it and the cost 
in each of the following examples by the letters of the 
word manuscript , if the cost is : 

1. $ .86, to be sold at a profit of: a. 15 % ; b. 20 %. 

2. $1.13, to be sold at a profit of: a. 20 % ; b. 121 %. 

3. $7.83, to be sold at a profit of: a. 25 % ; b. 30 %. 

4. $ .68, to be sold at a profit of: a. 35 % ; b. 25 %. 

5. $ 1.42, to be sold at a profit of: a. 33^ % ; b. 20 %. 

6. $ 5.35, to be sold at a profit of: a. 10 % ; b. 121 %. 

7. $ 12.40, to be sold at a profit of : a. 16f % ; b. 15 %. 

8. $ 26.25, to be sold at a profit of: a. 121 % ; b. 10%. 

9. $32.15, to be sold at a profit of: a. 20 % ; b. 25 %. 

10 . $ 65.35, to be sold at a profit of: a. 40 % ; b. 371 %. 

11 . A merchant bought: 

| dz. # 200 Aluminum Teapots @ $3.00 each, less 15 Jo. 
i dz. #400 Aluminum Teapots @ $3.50 each, less 15 °Jo . 
i dz. ft 173 Chafing Dishes @ $ 8.50 each, less 25 °/o. 

a. Hind the net cost of these articles, if purchased and 
paid for so as to receive a discount of 2 % for cash. 

b. Determine the selling price for each article, if sold so 
as to make a gross profit of 40 % over the net cost. 

12 . A merchant bought the following goods : 

1 doz. horse blankets, #3721, size 84 x 90 @ $8.00, less 45 °fo . 

£ doz. horse blankets, #3945, size 84 x 90, @ $9.50, less 40 Jo. 

1 doz. rawhide halters, # 592, 1| in., @ $45.00 per dozen, less 
30 Jo. 

a. Determine the total net cost of these articles. 

b. Determine the selling price for each single article so 
as to gain 20 % on the net cost. 


BUYING AND SELLING GOODS 


239 


Commission and Brokerage 

161. Firms and individuals often buy or sell goods or 
conduct other business transactions through agents. Such 
agents are also called Salesmen, Commission Merchants, or 
Brokers. 

A Commission Merchant buys and sells farm, garden, and 
dairy produce, and groceries. 

A Broker arranges business bargains and contracts be¬ 
tween parties. There are real estate brokers, stock brokers, 
bond brokers, cotton brokers, etc. 

The person or persons for whom an agent acts is called 
his Principal. 

The fee charged by an agent in general is called a Com¬ 
mission. That charged by a broker is called Brokerage. In 
many cases the fee is a certain per cent of the money value 
of the property involved in the transaction. 

162. Goods shipped by a commission merchant to his 
principal or by a principal to his agent is called a Shipment; 
goods received by either from the other is called a Consign¬ 
ment. 

A commission merchant, after selling goods for a principal, 
sends him an Account Sales ; after buying goods for a prin¬ 
cipal, he sends him an Account Purchases. 

The Gross Proceeds of a sale is the sum received from the 
purchaser. The Net Proceeds is the sum remaining for the 
principal after all charges have been deducted. 

In case a commission merchant purchases goods for'a 
principal, the Prime Cost is the actual sum paid for the 
goods. The Gross Cost is the sum of the prime cost and all 
other expenses involved in making the purchase and ac¬ 
quiring possession of the goods, such as freight, cartage, etc. 

163. The commission fee, when expressed as a per cent, is 
based upon the gross proceeds of a sale or on the prime cost of a 
purchase. 


240 


NEW HIGH SCHOOL ARITHMETIC 


For the account of 


An Account Sales 

^ 7 26 , f<73d 

Milwaukee , Wis .,- 

M. B. Caldwell & Company 
Commission Merchants 

jovcLou'W, ?MaAAA,la,n,cL, ZO-Ub. 


Date 

Rec’d 

Cab 

Contents 

Date 

Sold 

Charges 

Quantity 

Grade 

Price 

Amount 

3/12 

37592 

Rye 

25250 lb. 
Wheat 
54920 lb. 

3/12 



150.89 bu. 

915.33 bu. 
45.76 

3 

2 R.S. 

5 </o 

$2.83 

1276 

02 







869.57 


2.15 

1869 

57 










3145 

59 

Freight. 

Switching. 

Weighing and Inspection . 
Commission (1£ 0 per bu.) 

$70 

2 

19 

18 

00 

80 




91 

98 

Net proceeds 

3053 

61 


Note. — Observe that the weight of rye and wheat are given in 
pounds and are converted into bushels. See Tables for the weights 
of each per bushel. 

Also observe that the wheat suffers a reduction because of im¬ 
purities. 2 R. S. means # 2 Red Spring Wheat. 

EXERCISE 105 

1. What are the net proceeds to a farmer whose agent 
sells for him 1975 bn. of corn, at $ 1.331 per bushel, charg¬ 
ing 2 % for commission and $ 47.50 for freight and switch¬ 
ing charges ? 



































BUYING AND SELLING GOODS 


241 


2. An agent sold 3500 yd. of silk at $2.18 per yard. 
His charges were: commission, 21 °j 0 ; freight, $ 135.00. 
What were the net proceeds ? 

3 . What were the net proceeds on the sale of 375 bbl. of 
flour at $ 10.40 per barrel, if the commission was 3^ %, and 
charges for freight and storage were $ .67 per barrel ? 

4 . An attorney charged 10 % for collecting a debt of 
$ 945.00. He submitted also incidental charges of $ 15.00. 
What were the proceeds to his client ? 

5 . What are the net proceeds of a transaction if the 
gross proceeds are $ 786.45, the commission rate being 

%, and other charges $ 22.65 ? 

6. What rate of commission is an agent charging if the 
gross proceeds of a transaction are $ 387.50 and the com¬ 
mission is $ 9.69 ? 

7 . A college student sold during eight weeks of his sum¬ 
mer vacation 730 copies of a book at $ 1.50 each, receiving 
a commission of 35 His expenses averaged $ 15.00 per 
week. What did he make during his trip ? 

8. A man bought 1500 bu. of oats at $ .69 per bushel, 
paying a commission fee of 1 %, and other charges of 
$ 15.00. A few days later, he sold them for $ .73J- per 
bushel, again paying a commission of 1 °Jo> What did he 
gain or lose by the transaction ? 

9. The gross proceeds of a transaction are $2748.00 
There are charges of $12.50. The net proceeds are 
$2598.10. What rate of commission (on gross proceeds) 
was charged? 

10. An attorney charges $ 50.00 for his services in col¬ 
lecting a debt of $ 725.00. What rate of commission did 
he receive ? (Carry result to only one decimal place.) 


242 


NEW HIGH SCHOOL ARITHMETIC 


Use this year in dates. 

11 . Rule and fill in an account sales from the following 
data: William Parker and Son, Commission Merchants, 
Chicago, Ill., under date of Oct. 15, notified W. J. Brown, 
Park Ridge, Ill., that they had received on Oct. 11 a car, 
No. 44620, of barley, weighing 47,520 lb. ; that they sold 
it on Oct. 13, as grade 4 barley, at $ 1.35 per bushel. 
Charges : freight, $ 39.59 ; weighing, $ .55 ; inspection, 
$ 1.20; commission, 1} $ per bushel. 

12. Rule and fill in an account sales from the following 
data: J. C. Benedict, St. Paul, Minn., on Sept. 28, noti¬ 
fied Henry Sanstrom, Big Palls, Minn., of the receipt on 
Sept. 22 of car No. 60538, containing 50,350 lb. of wheat, 
containing 5 % impurities; that they sold it on Sept. 22 
at $ 2.18 per bushel. Charges : freight, $ 39.45 ; weighing, 
$ .80; inspection, $ 1.35 ; commission, 1} ^ per bushel. 

13 . James Barton, Des Moines, Iowa, sold for Oscar 
Sorenson, Fairfield, Iowa, 27-} T. of hay at $ 23.75 per ton. 
He charged a commission of 3 %. He paid freight of 
$42.85. Render an account sales, under date of Nov. 22. 

14 . Render to the Midvale Market Co., Midvale, Penn., 
an account purchase of the following articles, bought for 
them on June 21, by M. L. Selden, Fruit and Produce 
Merchant of Pittsburgh, Penn. : 

3 tubs containing 50 lb. each of butter, at $ .47 per pound; 
10 cases, containing 30 doz. each of eggs, at $ .34 per dozen; 
5 bunches of bananas, at $ 3.25 per bunch; 2 boxes oranges 
at $ 7.75 per box ; 10 crates of peaches, at $2.50 per crate; 
2 crates of pineapple, at $4.75 per crate; 5 boxes of Michi¬ 
gan celery, at $2.75 per box; 500 lb. of new potatoes, at 
$ 2.85 per hundred. Charges : commission, 5 % ; cartage, 
$4.25; freight, $6.35. 


BUYING AND SELLING GOODS 


243 


15 . R. L. Burton and Co., Chicago, Ill., sold for the 
Union Grove Creamery, Union Grove, Ill., the following 
consignments of butter during one month, charging a com¬ 
mission of 2 %. Make out an account sales, under date of 
June 30. 

June 8, 15 tubs, averaging 50 lb., at $.43 per pound; 
freight, $ 1.62. 

June 18, 20 tubs, averaging 50 lb., at $.45 per pound; 
freight, $2.28. 

June 24, 13 tubs, averaging 50 lb., at $ .44 per pound; 
freight, $ 1.52. 

J une 29, 12 tubs, averaging 50 lb., at $ .46 per pound; 
freight, $ 1.33. 

Cartage on each consignment. $ .10 per tub. 

16 . A broker buys for his principal, securities valued at 
$ 9500.00, charging brokerage of i %. What was the 
brokerage ? 

17 . The Central Realty Company, Melrose, Conn., sold 
for Hannah Putnam a farm of 84 A., at $ 72.50 per acre. 
The charges were: for bringing down to date the abstract 
of title, $3.75; incidental expenses, $17.75; commission, 
5 %. What were the net proceeds ? 

18. Barker and Davis, Auctioneers, charge a commission 
of 5 % for their services. What are the net proceeds to the 
owner of a farm if Barker and Davis sell at auction their 
farm personal property for $ 3486.38 ? 

The following problems, while not practical, sometimes 
appear in examinations. 

19. An agent received $ 2500 to invest, out of which he 
must receive enough to pay him his own commission at 
21%. How much can he invest and what will be his 
commission ? 

Suggestion. Since the commission is computed on the prime 
cost, then $ 2500 is 102£ °}o of the prime cost of the goods pur¬ 
chased. 


244 


NEW HIGH SCHOOL ARITHMETIC 


20. A merchant received $ 2194.50, with which to pur- 
chase flour at S 9.37^ per barrel, and pay him his own com¬ 
mission at the rate of 41%. How many barrels of flour 
should he purchase ? 

21. An agent charges a commission of 4 % for purchas¬ 
ing goods. If his commission was $ 352.60, what was the 
prime cost of the goods purchased ? 

22. What sum must be sent to a broker to enable him 
to buy 15 T. of hay at $ 23 per ton, and pay his commis¬ 
sion at the rate of 3i % ? 

23 . A commission merchant purchases for a principal 
goods at a prime cost of $ 2384.65. His fee is at the rate 
of 21 % ; other charges are $ 73.42. What was the gross 
cost of the shipment ? 

24 . An agent sold for a grain dealer a carload of wheat, 
containing 875 bu. of wheat. He sent the dealer an account 
sales, showing gross proceeds of $ 1006.25, and net proceeds 
of $ 965.58. If the miscellaneous charges were $ 27.54, what 
was the charge per bushel for selling the wheat ? 

25 . An agent sold goods amounting to $ 9220. He sub¬ 
mitted a bill for $ 404.25, including % 58.50 for freight and 
storage, and the balance for his commission. What was 
his rate of commission ? 


IX. INTEREST 


164. Interest is money paid for tlie temporary use of 
some other person’s money. It is like rent paid for the 
temporary nse of a house, farm, boat, or any other piece of 
property belonging to another person. 

The Principal is the money borrowed, on which interest is 
paid. The Amount is the sum of the principal and interest. 

Interest is usually described and computed as a certain 
rate per cent of the principal for one year. This rate per 
cent is the Rate of Interest. It will be understood that the 
rate is for one year. The year is usually regarded as con¬ 
sisting of 12 months of 30 days each or 360 days. 

165. Simple Interest is interest computed upon the prin¬ 
cipal alone. 

166. General Method of Computing Interest at Any Rate. 

Example 1. Find the interest on $ 2000 at 5 % for 
3 years. 

Solution. 1. The interest for 1 yr. = .05 X §2000 

= § 100 . 

2. .•. the interest for 3 yr. = 3 x §100 

= §300. 

Example 2. Find the interest on $850 for 2 yr. and 
7 mo. at 7 %. 

Solution. 1. The interest for 1 yr. = .07 X §850 

= §59.50. 

2. .-. the interest for 2 yr. = 2 X §59.50 

= §119.00. 

3. The interest for 1 mo. = T V of §59.50 = §4.958. 

4. .*. the interest for 7 mo. = 7 X §4.958 = §34.71. 

5. .*. the total interest = §119.00 + §34.71 = §153.71. 

245 


246 


NEW HIGH SCHOOL ARITHMETIC 


Example 3. Find the amount of $775 for 7 mo. and 
11 days at 5 %. 

Solution. 1. The interest for 1 yr. = .05 x $775 = $38.75. 

2. .*. the interest for 1 mo. = of $38.75 = $3,229. 

3. .*. the interest for 7 mo. = 7 x $3,229 = $22,603, or $22.60. 

4. Since 11 days = mo., the interest for 11 days = H x $3,229 

= ® 35 - 51 - = $1,183+ or $1.18. 

30 

5. .*. the total interest is $22.60 + $1.18 = $23.78. 

6 . .*. the amount = $775 + $23.78 = $798.78. 

Rule. — To find the interest on a given principal for a given 
time at a given rate, multiply the principal by the rate, and 
that product by the time expressed in years. 

Note 1. — In practice, one uses the method which best suits 
the problem. 

a. If the time is a number of years, multiply the interest for 
one year by the number of years. 

b. If the time involves months, find the interest for one month 
by dividing the annual interest by 12 , and then multiply by the 
number of months. 

However, for such a time as 8 mo., take T 8 ^ or § of the annual 
interest. 

Add the interest for the number of months to the interest for 
the given number of years. 

c. If the time involves days, consider 30 days as forming a 
month. Divide the monthly interest by 30, or the annual interest 
by 360, and multiply by the number of days. Add the result to 
the interest for the given number of years and months. 

However, if the time is such as 12 days, note that 12 days is 
or § of a month, hence the interest for 12 days is \ the monthly 
interest. Similarly, 12 days = °f a year, and hence the in¬ 
terest for 12 days is -jVo ° r 3 V of the annual interest. 

Note 2. — Carry out computations only to mills, and then in¬ 
crease the number of cents by 1 if the number of mills is 5 or more 
than 5. Thus, $29,637 will be called $29.64. Otherwise, ignore 
the number of mills. Thus, $29,633 will be called $29.63. 



INTEREST 


247 


EXERCISE 106 


Find the interest and the amount: 


Ex. No. 

Princ. 

Rate 

Time 

Ex. No. 

Princ. 

Rate 

Time 

1 . 

$2500 

4 </o 

3 yr. 9 mo. 

9 . 

$1500 

7 °!o 

3 mo. 10 da. 

2 . 

8300 

5% 

2 yr. 4 mo. 

10 . 

1800 

8 % 

5 mo. 9 da. 

3 . 

950 

6 % 

6 yr. 2 mo. 

11 . 

860 


2 yr. 4 mo. 10 da. 

4 . 

675 

8 % 

2 yr. 3 mo. 

12 . 

975 

4 \°lc 

3 yr. 2 mo. 20 da. 

5 . 

475 

5% 

1 yr. 5 mo. 

13 . 

1650 

H °Io 

4 yr. 3 mo. 9 da. 

6 . 

250 

5% 

8 mo. 15 da. 

14 . 

375 

4f °/o 

2 yr. 5 mo. 21 da. 

7 . 

1200 

6 % 

9 mo. 18 da. 

15 . 

1350 

6 |% 

1 yr. 7 mo. 22 da. 

8 . 

750 

4 °Jo 

7 mo. 12 da. 

16 . 

3275 

4 \°lo 

2 yr. 5 mo. 18 da. 


Note. — If more drill examples are wanted, find the interest on 
each of these sums at the rate and for the time given in Ex. 1, or 
in Ex. 2, etc. 

167. Six per cent is a commonly used rate of interest. 
This rate, combined with the assumed thirty-day month, 
leads to the use of the Six Per Cent Method of computing 
interest. 

The interest on any principal for a year at 6 % is .06 of 
the principal. Hence, the interest for two months is .01 of 
the principal and for one month is -i of .01 of the principal. 

Rule. — To find the interest at 6 % for any number of 
months: 

(1) Find .01 of the principal by pointing off two places. 

( 2 ) Then divide by 2 and multiply by the number of months. 

Note. — When the number of months is even, multiply the 
result of Step 1 by \ the number of months. 

Thus, the interest on $175.25 at 6 °fo for 10 months is 
5 x $1.7525 = $8.7625 = $8.76. 

And the interest on $175.25 at 6 °fo for 7 months is 

7 x \ of $1.7525 = 7 x $.87625 = $6.13375 = $6.13. 
















248 


NEW HIGH SCHOOL ARITHMETIC 


Since 6 days is -J- of the 30-day month, the interest for 6 
days is ^ of ^ of .01 or .001 of the principal. 

Rule. — To find the interest at 6 % for any number of days : 

(1) Find .001 of the principal by pointing off three places. 

( 2 ) Then divide by 6 and multiply by the number of days. 

Note. — If the number of days is divisible by 6, multiply the 
result of Step 1 by ^ the number of days. 

Thus, the interest on $350 at 6 °Jo for 18 days is -y~ 01 
3 x $ .350 = $ 1.05. 

And the interest on $ 350 at 6 °Jo for 13 days is 13 x 1 x $ .350 
= 13 x $.0583 = $.76. 


EXERCISE 107 


Find the interest at 6 % on each of the following sums 
of money for: 


a. 8 mo.; b. 

10 mo.; c. 2 

mo.; d. 5 mo.; e. 

7 mo.; /. 3 mo. 

1 . 

$360 

4 . $136 

7 . $425.50 

10 . $1632.50 

2 . 

$140 

5 . $185 

8 . $563.80 

11 . $2143.25 

3 . 

$268 

6 . $940 

9 . $147.65 

12 . $5063.75 


13 - 24 . Find the interest at 6 % on each of the foregoing 
sums for: 

a. 12 da. c. 42 da. e. 14 da. g. 15 da. 

b. 18 da. d. 54 da. /. 21 da. h. 9 da. 

25 . Find the interest at 6 % for 1 yr. 7 mo. and 20 da. on 
$2755. 

Solution. 

1. The interest for one year = $ 165.30 

2. The interest for 7 mo. = 96.43 ( = 7 x 1 x 27.55) 

3. The interest for 20 da. = 9.18 ( = 20 x £ x 2.755) 

4. the total interest =$270.91 



INTEREST 


249 


Find the interest at 6 % 

26 . $ 641 for 5 yr. 10 mo. 

27 . $ 71.25 for 2 yr. 5 mo. 

28 . $ 833 for 3 yr. 2 mo. 

29 . $ 10.25 for 1 yr. 4 mo. 

30 . $ 2575 for 4 yr. 7 mo. 


on: 

31 . $ 146.25 for 1 yr. 3 mo. 12 da. 

32 . $ 275.50 for 3 yr. 4 mo. 18 da. 

33 . $ 735.75 for 5 yr. 2 mo. 27 da. 

34 . $ 227.60 for 1 yr. 14 da. 

35 . $ 2450 for 1 yr. 7 mo. 11 da. 


168. Finding the Interest at any Rate by the Six Per Cent 
Method. 

Since 5 % is f of 6%, then the interest at 5 % is -J of the 
interest at 6 %. Hence, to find the interest at 5 °] 0 , 
multiply the interest at 6 % by f or subtract from the 
interest at 6 % one sixth of itself. 

% is three fourths of 6 %. Hence, to find the interest 
at 4^ %, subtract from the interest at 6 % one fourth of 
itself. 

Example. Find the interest on $ 1250 for 3 mo. and 
15 da. at: a. 5 % ; b. 4 % ; c. %. 

Solution. g 

1. The interest on $ 1250 at 6 °jo for 3 mo. = ^ x $1£.^0 = $ 18.75. 

r 

j S .625 

2. The interest on $ 1250 at 6 °lo for 15 da. =— X = $3,125 

P 

% 

3. .•. the interest on $1250 at 6 °Jo for 3 mo. 15 da. = $21,875. 

_ 3.645+ 

4 . a. The interest at 5 °/o = - x $ 

= $18,225 = $18.23. 

b. The interest at 4 % = f the interest at 6 <Jo 

0 7.291+ 

= “ x $£f.W = $ 14.582 = $ 14.58. 

3 

c. The interest at 4£ °Jo = f the interest at 6 °Jo 

„ 5.468+ 

= _ x $ ?;.8jT£ = $ 16.404 = $ 16.40. 


250 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 108 


Find the interest by the 6 % method: 


Ex. No. 

On 

For 

At 

a 

b 

c 

d 

1 . 

$530 

8 mo. 

5% 

7 % 

8 % 

4 % 

2 . 

450 

5 mo. 

9 % 

3 % 

2 % 

l % 

3 . 

275 

4 mo. 18 da. 

5% 

8 % 

4 1% 

71% 

4 . 

1250 

3 mo. 15 da. 

3 % 

4% 

9 % 

5% 

5 . 

252.50 

1 yr. 5 mo. 

4 1% 

H% 

l 1 % 

8 % 

6. 

250 

75 da. 

5% 

7 <jo 

9% 

11% 

7 . 

375 

60 da. 

3 % 

4 % 

4 1% 

8 % 

8 . 

750 

35 da. 

7% 

9 % 

2 % 

41% 

9. 

1500 

40 da. 

5% 

4 % 

4 1% 

11% 

10 . 

500 

85 da. 

5 1% 

2 § % 

31% 

41% 


169. When money is loaned, the interest is usually pay¬ 
able semi-annually or quarterly. 

i Often men must borrow a sum of money for a few days, 
and do so at such rates as 3-f %, 5^ %, etc. 

Computations such as those in the following examples 
result. 

EXERCISE 109 


Find the semi-annual interest on : 


1. 

$750: 

a. at 6 % 

6. at 5% 

c. 

a f °Jo 

2. 

$2500: 

a. at 51 % 

5. at 6 % 

c. 

at 4J% 

3. 

$4000: 

ci. at 6-g- % 

5. at 5^ % 

c. 

at 4%. 

4. 

$500: 

a. at 41- % 

6. at 51 % 

c. 

at 6%. 

5. 

$825: 

a. at4f% 

5. at 51 °J 0 

c. 

at 6 %. 


Find the interest on : 


6. $2300 at 5|%: a. for 10 da. 5. for 12 da. c. 15 da. 

7. $ 1500 at % : a. for 9 da. b. for 13 da. c. 17 da. 















INTEREST 


251 


8. $675 at5J%: a. for 11 da. b. for 12 da. c. 8 da. 

9. $ 3750 at % : a. for 16 da. b. for 10 da. c. 20 da. 

10. $ 15,000 at 5J % : a. for 4 da. b. for 9 da. c. 15 da. 

170. If the time is less than one year, the number of 
months and days may be computed as in the following 
example. 

Example. A man bought $ 175 worth of coal on July 17, 
1932, agreeing to pay 6 % interest on this sum until the bill 
was paid. He paid it on Nov. 10, 1932. How much in¬ 
terest did he have to pay ? 

Solution. 1. From July 17 to Oct. 17 is three months; from 
Oct. 17 to Oct. 31 is 14 days; from Oct. 31 to Nov. 10 is 10 days. 
Hence the total time is 3 months 24 days. 

2. The interest on $175 at 6% for 3 mo. = $2.63. (§ 167) 

3. The interest on $175 at 6% for 24 days = $ .70. 

4. the total interest = $3.33. 

If the time is more than one year, find first the number 
of years, and then the number of months and days as in the 
following example. 

Example. Find the interest at 6 % on $1150 from 
June 18, 1931, to April 11, 1933. 

Solution. 1. From June 18, 1931, to June 18,1932, is 1 year. 

2. From June 18, 1932, to March 18, 1933, is 9 months. 

3. From March 18, 1933, to April 11, 1933, is 24 days. 

4. Hence, the time is 1 year, 9 months, 24 days. 

5. The interest on $1150 for 1 yr. at 6 °Jo is $ 69.00. 

6 . The interest on $1150 for 9 mo. at 6% is$ 51.75. 

7. The interest on $1150 for 24 da. at 6% is $ 4.60. 

8 . .-. the total interest is $125.35. 


252 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 110 


Find the interest: 


Ex. No. 

On 

At 

From This Year 

To Next Year 

1 . 

$685 

6% 

June 13 

Jan. 25 

2 . 

4175 

6% 

April 10 

May 12 

3. 

5160 

5% 

Feb. 21 

Mar. 14 

4. 

950 

4% 

Nov. 15 

July 26 

5. 

498 

H% 

July 14 

May 21 

6 . 

475 

5% 

Nov. 29 

Sept. 9 

7. 

550 

4i% 

Jan. 26 

Mar. 12 

8 . 

8780 

3% 

Apr. 4 

Nov. 17 

9. 

225 

5i% 

Dec. 22 

Feb. 3 

10 . 

175 

41% 

Apr. 6 

May 12 


171. Since the calendar year consists of 365 days, there 
is a slight inaccuracy in basing interest computations upon 
an assumed year of 360 days. 

Thus, the interest for 40 days should be of the annual 
interest instead of The interest therefore should be a little 

less than that charged by the computations based upon a 360-day 
year. 

Interest calculated by considering the exact number of 
days in the given time and using 365 days for the year is 
called Exact Interest. It is used by the United States Gov¬ 
ernment in computing interest on its securities. It is also 
used by some business houses. 

Example. Find the exact interest at 6 % on $ 2500 from 
June 12, 1931, to Sept. 8, 1933. 

Solution. 1. From June 12, 1931, to June 12, 1933, is 2 years. 

2. From June 12, 1933, to June 30, 1933, is 18 days. 

From June 30, 1933, to July 31, 1933, is 31 days. 

From July 31, 1933, to Aug. 31, 1933, is 31 days. 

From Aug. 31, 1933, to Sept. 8, 1933, is 8 days. 











INTEREST 


253 


3. .*. the total time is 2 years and 88 days. 

4. The interest on $ 2500 at 6 % for 2 years is $ 300. 

5. The interest on $ 2500 at 6 °Jo for 1 year is $ 150. 

the interest for 88 days is ^ x $ 150 = $ 36.16. 

6. .*. the total interest = $300 + $36.16 = $336.16. 

EXERCISE 111 

In the first three examples, find the interest by the six 
per cent method and also the exact interest; in the remain¬ 
ing examples find only the exact interest. 

Eind the exact interest on: 

1. $ 925 at 6 % for 3 yr. 110 da. 

2. $ 250 at 5 % for 2 yr. 200 da. 

3 . $ 550 at 4 % for 75 da. 

4 . $ 7150 at 5 % for 92 da» 

5 . $ 1230 at 6 % for 1 yr. 15 da. 

Eind the exact interest: 


Ex. No. 

On 

At 

From 

This Year 

To 

Next Year 

6 . 

$650 

6 °jo 

May 

5 

Aug. 12 

7. 

225 

5 % 

March 10 

Sept. 22 

8 . 

1500 

4 °fo 

July 

15 

June 10 

9. 

25,000 

4 \1 o 

Aug. 

18 

Dec. 2 

10 . 

12,500 

3 % 

Sept. 

16 

Dec. 30 


172. Finding the Rate when the Principal, Interest and 
Time are Given. 

Example. What rate of interest has a man received who 
gets a return of $41.76 on an investment of $ 384 for 2 yr. 
and 5 mo. ? 

Solution. 1. By the 6 °fo method, the interest at 1 °/o on $384 
for 2 yr. 5 mo. is $ 9.28. 

2. .*. the required rate is the number of times $ 9.28 is con¬ 
tained in $41.76 or 4|; i.e., the rate is 4£ $>. 












254 


NEW HIGH SCHOOL ARITHMETIC 


Rule.—To find the rate when the principal or amount, 
interest, and time are given: 

1. Find the interest on the principal at 1 % for the given 
time. 

2 . Divide the given interest by the result of Step i, thus 
obtaining the rate per cent. 

Note. — If the interest and the amount are given, first find the 
principal by subtracting the interest from the amount. 

EXERCISE 112 

At what rate per cent of interest will: 

1. $ 480 earn $ 56.00 in 2 yr. and 11 mo. ? 

2 . $ 6450 earn $ 123.65 in 4 mo. 18 da. ? 

3. $ 3780 earn $ 11.76 in 28 da. ? 

4 . $ 750 earn $ 34.38 in 10 mo. ? 

5. $ 225 earn $ 29.25 in 4 yr. 4 mo. ? 

6 . $ 8500 amount to $ 8780.50 in 8 mo. 24 da. ? 

7. $ 720 amount to $ 734.70 in 7 mo. ? 

8 . $ 192 amount to $ 202.56 in 1 yr. 10 mo. ? 

9. What rate of interest has a man earned on an invest¬ 
ment of $ 750 if he gets back $ 875 at the end of 15 mo. ? 

10 . A man can buy his coal for $ 105 cash, or $ 110 if he 
pays for it in 60 days’ time. What rate of interest does he 
pay if he chooses to pay in 60 days’ time ? 

11 . A home was purchased for $ 5100, and was sold two 
years later for $5800. The owner reckoned his rent as 
worth $ 300 per year to him. He paid out $ 40.00 per year 
for taxes, $35.00 per year for insurance and $65.00 per 
year for repairs. What per cent of income on his invest¬ 
ment did he make ? 

12 . A farm has a total value of $18,500. The net in¬ 
come for the year is $ 1250. What rate of interest is that 
on the investment ? 


INTEREST 


255 


13 . A man borrows $ 125 and at the end of 60 days re¬ 
turns to the lender $ 130. What rate of interest has he 
paid on the borrowed money? 

14 . A man lends $2500 to a company. During the year 
he receives $160 interest. What rate of interest is his 
money earning ? 

15 . From another investment of $ 725 he receives during 
the year an income of $ 60. What rate of interest is this ? 

173. Finding the Time when the Principal or Amount, In¬ 
terest, and Rate are Given. 

Example. How long must $ 1000 be on interest at 6 % 
to double itself ? 

Solution. 1. Since the money doubles itself, it earns $1000 in 
interest. 

2. The interest on the given principal, $1000, for 1 yr. at 6 °fo 
is $60.00. 

3. Therefore it will take as many years as $ 60 is contained times 
in $1000. 

4. This is 16 years, or 16 yr. and 8 mo. 

Rule. — To find the time when the principal, rate, and in¬ 
terest are known: 

1. Find the interest on the principal at the given rate for one 
year. 

2 . Divide the given interest by the result of Step i, thus ob¬ 
taining the number of years in the required time. 

3 . Express the time in years, months, and days. 

Note. — By carrying the division out to three decimal places, 
sufficiently accurate results are obtained. Thus, if the time should 
happen to be 2.736 yr., the years are 2. 

.736 yr. = .736 x 12 mo. = 8.832 mo. 

.832 mo. = .832 x 30 da. = 24.96 da. or 25 da. 

Hence the time is 2 yr. 8 mo. 25 da. 


256 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 113 

Find the time in which : 

1. $92 will gain $ 11.73 at 6 %. 

2. $ 250 will gain $ 85.00 at 6 %. 

3. $ 696 will amount to S 713.40 at 5 

4 . $ 637.50 will amount to $ 1020 at 4 

5. $ 936 will gain $ 54.60 at 3 %. 

6. $ 295 will gain $ 17.10 at 4 %. 

7 . $ 1060 will amount to $ 1200 at 5 

8 . $ 716.84 will gain $ 42.40 at 6 %. 

9. $476.50 will amount to $489.20 at 4 

10 . $ 1816 will gain $ 19.08 at 6 %. 

11 . $ 800 will amount to $ 1000 at 4 1 %. 

12 . $ 265 will amount to $ 500 at 5 %. 

13 . $ 57 will gain $ 13 at 3 %. 

14 . In what time will $ 500 double itself at 5 % ? 

15 . In what time will $ 800 double itself at 5 % ? 

16 . Judging from Examples 14 and 15, in what time will 
any sum double itself at 5 % ? 

Verify your conclusion by testing it for some definite 
sum. 

17 . In what time will any sum double itself at 4 % ? 

174. Finding the Principal when the Interest or Amount, 
Time, and Rate are Given. 

Example. What principal at 6 % will amount to $ 3872 
in 3 yr. 6 mo. ? 

Solution. 1. The interest on $ 1 for 3 yr. 6 mo. is $ .21. 

Hence $ 1 will amount to $ 1.21 in 3 yr. 6 mo. 

2. it will take as many dollars to amount to $3872 as $1.21 
is contained times in $ 3872. 

3. This is $3200. 


INTEREST 


257 


Rule. — To find the principal when the rate, time, and inter¬ 
est are given: 

1. Find the interest on $1 for the given time at the given 
rate. 

2 . Then divide the given interest by the result of Step i. 
This will give the principal. 

3 . Check the result. 

Note. —If the amount is given, find the amount of $ 1.00 at the 
given rate for the given time. Divide the given amount by this 
result to get the principal. 

EXERCISE 114 

What principal at simple interest will: 

1. Gain $ 136.40 in 3 yr. 8 mo. at 6 % ? 

2 . Gain $ 13.82 in 1 yr. 6 mo. at 8 % ? 

3 . Amount to $ 503.97 in 9 yr. 6 mo. at 6 % ? 

4 . Gain $ 22.29 in 10 mo. at 3 % ? 

5. Amount to $ 431.20 in 2 yr. 6 mo. at 4 % ? 

6. Gain $ 132.00 in 1 yr. 3 mo. at 4 % ? 

7. Amount to $ 50.14 in 1 yr. 1 mo. 15 da. at 8 % ? 

8. Gain $ 109.62 in 11 mo. 18 da. at 6 % ? 

9. Amount to $ 385.00 in 9 mo. 6 da. at 7 \°Jo c ! 

10. Amount to $ 921.93 in 2 mo. 12 da. at 6 °Jo ? 

11. Gain $ 42.25 in 1 yr. 1 mo. at 6 % ? 

12. Amount to $ 2515.35 in 4 mo. 15 da. at 6 °Jo ? 

13 . Gain $ 9.60 in 12 da. at 3 % ? 

14 . Amount to $ 3556.91 in 7 mo. 12 da. at 6 % ? 

15 . Gain $ 44.85 in 4 mo. 18 da. at 4J % ? 


258 


NEW HIGH SCHOOL ARITHMETIC 


Compound Interest 

175. In some interest computations, the interest earned 
on the principal during a period of time like three months, 
six months, or a year is added to the principal and, with it , 
draws interest during the next period of time. 

When the interest is added periodically to the principal 
in this manner, the interest computed on the principal and 
the unpaid interest is called Compound Interest. In most 
cases, interest is compounded quarterly , semi-annually or an¬ 
nually. 

Example. Find the compound interest and the amount 
on $ 500 invested for 2 yr., at 6 % interest, compounded 
semi-annually. 

Solution. 

1. The interest for the first six months = 3 °/o of $ 500 = $ 15.00. 

2. .*. the amount at the end of 6 mo. = $ 515.00. 

3. The interest during the 2d 6 mo. = 3% of $515.00 = $ 15.45. 

4. .*. the amount at the end of 1 yr. = $ 530.45. 

5. The interest during the 3d 6 mo. = 3% of $ 530.45 = $ 15.91. 

6. .*. the amount at the end of 1^ yr. = $ 546.36. 

7. The interest during the 4th 6 mo. = 3 °Jo of $ 546.36 = $ 16.39. 

8. .*. the amount at the end of 2 yr. = $ 562.75. 

9. .•. the compound interest =$ 562.75 — $ 500 = $ 62.75. 

Note 1. — The simple interest at 6 % on $ 500 for 2 yr. = $ 60. 

Note 2. — In the following list of examples, carry out all the 
computations to four decimal places. 

EXERCISE 115 

1. Find the amount to which $412 will accumulate in 
5 years if it was invested on Jan. 2, 1933, at 4 % interest 
compounded quarterly. 

2 . What is the compound interest on $ 850 for 3 years 
at 6 %, compounded annually ? 


INTEREST 


259 


3. What is the amount to which $ 500 accumulates in 4 
years at 5 %, compounded annually? 

4 . What is the compound interest on $675 for 5 years 
at 4 %, compounded annually? 

5. What is the compound interest on $ 1050 for 3 years 
6 months at 6 %, the interest being compounded semi-annu¬ 
ally? 

6. What is the amount of $5000 for 2 years at 8%, 
compounded quarterly ? 

7 . What is the amount of $ 3000 for 2 years 6 months 
at 4 % interest, compounded semi-annually ? 

8. a. What is the compound interest on $ 6000 for five 
years at 4 %, compounded quarterly ? 

b. What is the simple interest on $ 6000 for five years 
at 4 % ? 

9. What is the amount of $1200 for 3 years at 4%, 
compounded semi-annually ? 

10 . What is the amount of $1200 for 3 years at 4%, 
compounded quarterly? 

Savings Bank Accounts 

176. A Savings Bank receives small sums of money for 
deposit, paying interest therefor. Interest is computed, as 
a rule, quarterly or semi-annually on certain specified 
Interest Days. The time between two successive interest 
days is called the Interest Period. Interest is allowed 
usually only on the smallest cash balance of the depositor 
during the interest period ending on the day interest is 
computed. Interest is not allowed on any fractional part 
of a dollar. 

Thus, if the smallest balance during a period is $273.67, 
interest is computed on $273. 


260 


NEW HIGH SCHOOL ARITHMETIC 


Interest is usually at the rate of 3 % or 4 %. 

If the interest is not drawn when payable, it is added to 
the principal, and draws interest as a new deposit; in this 
way savings banks pay compound interest. 


The following form of record is kept in a savings bank. 
It is assumed that interest at 4 % is computed on the 1st of 
January, April, July, and October. 


People’s Savings Bank 
In Account with Mr. Charles Goss 


Date 

Deposits 

Withdrawals 

Interest 

Balance 

1932 


$ 


$ 


$ 


$ 


Feb. 

2 

25 

00 





25 

00 

Mar. 

28 

100 

75 





125 

75 

Apr. 

1 







125 

75 

Apr. 

10 

75 

00 





200 

75 

May 

15 

50 

00 





250 

75 

July 

1 





1 

25 

252 

00 

Aug. 

15 

20 

00 





272 

00 

Sept. 

1 



75 

00 



197 

00 

Sept. 

24 

150 

00 





347 

00 

Oct. 

1 





1 

97 

348 

97 


This record shows that Mr. Goss opened his account on Feb. 2, 
1932. On April 1, he did not receive any interest since his 
smallest balance between Jan. 1 and April 1 is zero. On July 1, 
he received $ 1.25 interest since his smallest balance between April 
1 and July 1 was $125.75 and the interest on $125 for 3 months 
at 4 % is $1.25. He withdrew $75.00 on Sept. 1, and deposited 
$ 150 on Sept. 24. On Oct. 1, he received $ 1.97 interest, since his 
lowest balance was $ 197 on Sept. 1, and the interest on that 
amount for 3 months at 4 °Jo is $ 1.97. 
























INTEREST 


261 


EXERCISE 116 

Solve the following examples. Assume that the interest 
days are the 1st of January, April, July, and October. 

1. The South Side Savings Bank pays 4 % interest. 
Make out an account for Clara Danville. 

Deposits : April 1, $ 50; April 12, $ 15.25 ; April 30, 
$22.00; May 11, $16.00; June 20, $25.00; July 

16, $50.00; Aug. 5, $18.00; Aug. 27, $13.75; Sept. 19, 
$33.42; Oct. 10, $15.00; Oct. 23, $19.00; Nov. 16, 
$17.35; Nov. 22, $43.62; Dec. 3, $35.00. Withdrawals: 
Dec. 22, $75.00. Find balance on Jan. 1. 

2. The Mapleville Savings Bank pays compound interest 
at the rate of 3 %. Make out an account for George 
Kendall, and balance it on Feb. 16, 1933. (Deposits are 
marked +, and withdrawals -). Dec. 27, 1931, +$ 200; 
Jan. 22, 1932, +$85.32; Feb. 2, -$10.00; March 13, 
+$62.75; April 6, -$12.00; April 13, +$25.64; May 10, 
+$48.52; May 17, -$15.00; June 23, +$36.00; July 
2, -$25.00; July 19, +$7.82; Aug. 6, -$19.00; Sept. 5, 
+ $20.00; Sept. 17, +$24.00; Sept. 29,-$63.00; Oct. 14, 
-$17.00; Oct. 26, +$46.85; Nov. 18, +$73; Dec. 20, 
-$50.00. 

3. The Westport Savings Bank pays 4 % interest. 
Make out an account for John Clews down to Jan. 5, 1933. 

Feb. 16, 1932, +$55; Feb. 26, +$32.50; Mar. 19, 
+ $27.85; Apr. 10, +$83; Apr. 17, -$25; Apr. 28, 
+ $63.75; May 14, +$95; June 6, +$120.36; June 28, 
+ $ 165.22 ; July 6, +$ 85 ; July 24, -$ 50 ; Aug. 19, +$ 16 ; 
Aug. 26, +$20; Sept. 4, +$32.50; Sept. 16, +$29.45; 
Sept. 29, +$63.75; Oct. 6, -$35; Oct. 15, +$13.50; Oct. 
24, +$26.72; Oct. 31, +$16.48; Nov. 18, -$15.00; Nov. 
22, +$11.75; Nov. 30, +$35.25; Dec. 10, +$64; Dec. 15, 
+ $37.00; Dec. 20, +$12.50; Dec. 28, +$125; Jan. 5, 1933, 
-$ 20 . 


262 


NEW HIGH SCHOOL ARITHMETIC 


SUPPLEMENTARY TOPIC 

177. Postal Savings Banks provide a convenient and safe 
means of saving and investing small sums of money. 

Any person 10 years of age or over may deposit money 
in a postal savings account in the nearest post office in 
which savings accounts are kept. 

For amounts less than $ 1, the depositor buys postal sav¬ 
ings stamps at 10^ each. These are affixed to a card. 
After ten stamps are affixed to the card, the $ 1 represented 
may be used to start a postal savings account or may be 
deposited in an account already established. Sums of $ 1 
or more may be deposited directly. 

Postal Savings certificates, valued at $ 1, $ 2, $ 5, $ 10, 
$20, $50, $100, $200, $500, may be purchased in the 
above manner. The largest amount any person may deposit 
is $ 2500. 

The sums deposited draw interest at the rate of 2 % per 
year. The interest is computed only on whole dollars — 
not upon fractional parts of a dollar. 

A depositor may withdraw his deposits at any time at 
the post office through which he has made his deposits; or 
he may transfer his deposits from one post office to another 
without losing any interest, and without any cost. 

After a depositor has as much as $ 20, $ 100, or $ 500, he 
may, after 30 days’ notice, exchange his certificates of de¬ 
posit for United States Bonds which bear 2J %, payable 
semi-annually. 


EXERCISE 117 

1. John receives an allowance of 50^ per week. He 
decides to place 25 $ of it each week in the postal savings 
bank. 

How much will John have in the bank at the end of a 
half year ? 


INTEREST 


263 


2. Continue the following table for a period of 10 years : 


John’s Postal Savings Account 


Dates 

Balance 

Interest 

Deposits 

Total 

1/1/33-7/1/33 

$00.00 

$0.00 

$6.50 

$ 6.50 

7/1/33-1/1/34 

6.50 

.06 

6.50 

13.06 

1/1/34-7/1/34 

13.06 

.13 

6.50 

19.69 


3. How much of John’s balance at the end of 10 years is 
interest on his deposits ? 

178. In practice, the solution of problems in compound 
interest is abridged by the use of a table like that on p. 265. 
In this table is given the amount of $1.00 compounded 
annually for any number of years from 1 to 40, for certain 
rates per cent. 

Example. What is the amount of $ 375 in 5 yr. at 4 %, 
compounded annually ? 

Solution. 1. The amount of $1.00 in 5 yr. at 4</o, compounded 
annually, is $1.216653. 

2. .*. the amount of $375 is 375 x $1.216653 or $456.24. 

179. The complete usefulness of such a table depends 
upon a fact brought out by the following problems. 


I 


Find the amount of $ 1.00 at 1 %, compounded annually 


for four years. 

1. Interest during 1st yr. 

2. Amount end of 1st yr. 

3. Interest during 2d yr. 

4. Amount end of 2d yr. 

5. Interest during 3d yr. 

6. Amount end of 3d yr. 

7. Interest during 4th yr. 

8. Amount end of 4th yr. 


= .01 
= 1.01 
= .0101 
= 1.0201 
= .010201 
= 1.030301 
= .01030301 
= 1.04060401 











264 


NEW HIGH SCHOOL ARITHMETIC 


II 

Hind the amount of $ 1.00 at 4 compounded quarterly 
for one year. 


1 . 

2 . 

3. 

4. 

5. 

6 . 

7. 

8 . 


Interest during 1st quarter = 
Amount end 1st quarter = 
Interest during 2d quarter == 
Amount end 2d quarter = 
Interest during 3d quarter = 
Amount end 3d quarter = 
Interest during 4th quarter = 
Amount end 4th quarter = 


.01 

1.01 

.0101 

1.0201 

.010201 

1.030301 

.01030301 

1.04060401 


Examination of these solutions suggests the 


Rule. — i. The compound interest on and the amount of $ 1.00 
at a certain rate compounded semi-annually for a certain number 
of years is the same as the compound interest on and the 
amount of $ 1.00 at one half that rate compounded annually for 
double the number of years. 

2 . The compound interest on and the amount of $1.00 at a 
certain rate compounded quarterly for a certain number of years 
is the same as the compound interest on and the amount of 
$ 1.00 at one fourth that rate compounded annually for 4 times 
the number of years. 


Example. Find the amount and the compound interest 
on $ 250, at 6 %, compounded quarterly for 5 yr. 

Solution. 1. Since the interest is compounded quarterly, find 
the amount and the compound interest on $ 1.00 at \ of 6 °/o or 
1 \°Jo compounded annually for 4 x 5 or 20 yr. 

2. By the table, the amount is $ 1.346855 (page 265, col. 4). 

3. .-. by Rule 2, the amount of $ 250=250 x 1.346855 = $ 336.71+ 

4. .*. the compound interest = $336.71 — $250 = $ 86.71. 


INTEREST 


265 


Compound Interest Table 

Giving the Amount of $ 1 Compounded Annually at: 


No. 

1% 

1% 

H% 

2% 

3% 

4% 

1 

1.007500 

1.010000 

1.015000 

1.020000 

1.030000 

1.040000 

2 

1.015056 

1.020100 

1.030225 

1.040400 

1.060900 

1.081600 

3 

1.022669 

1.030301 

1.045678 

1.061208 

1.092727 

1.124864 

4 

1.030339 

1.040604 

1.061364 

1.032432 

1.125509 

1.169859 

5 

1.038067 

1.051010 

1.077284 

1.104081 

1.159274 

1.216653 

6 

1.045852 

1.061520 

1.093443 

1.126162 

1.194052 

1.265319 

7 

1.053696 

1.072135 

1.109845 

1.148686 

1.229874 

1 315932 

8 

1.061599 

1.082857 

1.126493 

1.171659 

1.266770 

1.368569 

9 

1.069561 

1.093685 

1.143390 

1.195093 

1.304773 

1.423312 

10 

1.077583 

1.104622 

1.160541 

1.218994 

1.343916 

1.480244 

11 

1.085664 

1.115668 

1.177949 

1.243374 

1.384234 

1.539454 

12 

1.093807 

1.126825 

1.195618 

1.268242 

1.425761 

1.601032 

13 

1.102010 

1.138093 

1.213552 

1.293607 

1.468534 

1.665074 

14 

1.110276 

1.149474 

1.231756 

1.319479 

1.512590 

1.731676 

15 

1.118603 

1.160969 

1.250232 

1.345868 

1.557967 

1.800944 

16 

1.126992 

1.172579 

1.268986 

1.372786 

1.604706 

1.872981 

17 

1.135445 

1.184304 

1.238020 

1.400241 

1.652848 

1.947901 

18 

1.143960 

1.196147 

1.307341 

1.428246 

1.702433 

2.025817 

19 

1-152540 

1.208109 

1.326951 

1.456811 

1.753506 

2.106849 

20 

1.161184 

1.220190 

1.346855 

1.485947 

1.806111 

2.191123 

21 

1.169893 

1.232392 

1.367058 

1.515666 

1.860295 

2.278768 

22 

1.178667 

1.244716 

1.387564 

1.545980 

1.916103 

2.369919 

23 

1.187507 

1.257163 

1.408377 

1.576899 

1.973587 

2.464716 

24 

1.196414 

1.269735 

1.429503 

1.608437 

2.032794 

2.563304 

25 

1.205387 

1.282432 

1.450945 

1.640606 

2.093778 

2.665836 

26 

1.214427 

1.295256 

1.472710 

1.673418 

2.156591 

2.772470 

27 

1.223535 

1.308209 

1.494800 

1.706886 

2.221289 

2.883369 

28 

1.232712 

1.321291 

1.517222 

1.741024 

2.287928 

2.998703 

29 

1.241957 

1.334504 

1.539981 

1.775845 

2.356566 

3.118651 

30 

1.251272 

1.347849 

1.563080 

1.811362 

2.427262 

3.243398 

31 

1.260656 

1.361327 

1.586526 

1.847589 

2.500080 

3.373133 

32 

1.270111 

1.374941 

1.610324 

1.884541 

2.575083 

3.508059 

33 

1.279637 

1.388690 

1.634479 

1,922231 

2.652335 

3.648381 

34 

1.289234 

1.402577 

1.658996 

1.960676 

2.731905 

3.794316 

35 

1.298904 

1.416603 

1.683881 

1.998890 

2.813862 

3.946089 

36 

1.308645 

1.430769 

1.709140 

2.039887 

2.898278 

4.103933 

37 

1.318460 

1.445076 

1.734777 

2.080685 

2.985227 

4.268090 

38 

1.328349 

1.459527 

1.760798 

2.122299 

3.074783 

4.438813 

39 

1.338311 

1.474123 

1.787210 

2.164745 

3.167027 

4.616366 

40 

1.348349 

1.488864 

1.814018 

2.208040 

3.262038 

4 801021 






















266 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 118 

Find the amount when compounded: a. annually; 
b. semi-annually; c. quarterly. 

1. Of $ 100 for 6 yr. at 3 %. 

2. Of $ 1000 for 5 yr. at 4 %. 

3 . Of $ 250 for 8 yr. at 4 %. 

4 . Of $ 350 for 10 yr. at 4 %. 

5 . A man made a bequest of $ 25,000 to a college, with 
the condition that it should be invested for 10 yr. in se¬ 
curities, paying 4 % interest, compounded quarterly. What 
will be the amount at the end of that time ? 

6 . Suppose that a savings bank pays 4 % interest quar¬ 
terly on January, April, July, and October 1st. Suppose 
that $ 1.00 is placed in the bank at least three days before 
each interest payment date, starting with January, 1928, 
and continuing until January, 1933. 

a. Make a table showing in one column the date of the 
deposits, in a second column, the value of each deposit, and 
in a third column, the amount to which each deposit ac¬ 
cumulates at compound interest by January 2, 1933. 

b. Find the total amount for all the deposits. 

c. Suppose that the quarterly deposit had been $ 5 in¬ 
stead of $ 1. Find the amount of all such deposits by 
multiplying the result of part b by 5. 

7. a. Obtain as in Example 6, the amount to which 
$ 1.00 per quarter will accumulate in ten years at 4 % in¬ 
terest, compounded quarterly, each deposit being made before 
the interest date, until forty deposits have been made. 

b. Using the result of part a, find the amount if the 
deposits had been $ 10.00 per quarter instead of $ 1.00. 

8 . In 1928 a boy was promised a Christmas gift of $ 50 
in each of the ten years from 1928 to 1937 inclusive, pro- 


INTEREST 


267 


vided he deposited it in a savings bank within three days 
and allowed it to accumulate at 4 % interest, compounded 
quarterly until Oct. 1, 1938, at which time he could begin 
drawing upon it for college expenses. How much money 
will be available at that time ? 

Suggestion. — This example, like Ex. 6 and Ex. 7, can be sim¬ 
plified by finding first the amount if the deposit were only $ 1.00 
and then multiplying this amount by 50. 

9 . A girl’s birthday is March 10th. In 1932 she is 9 
years old. She is promised a gift on each birthday, begin¬ 
ning with 1932, until she is 18 years old, of as many times 
$ 5 as she is years of age, provided she will agree to deposit 
the money within three days in a savings bank paying 4 % 
interest, compounded quarterly. What will be the amount 
of her account on Oct. 1 of the year in which she is 18 
years old ? 

10. A man and his wife, desiring to provide for their 
later years, decide to deposit $ 50 per quarter in a savings 
bank paying 4 % interest quarterly. Assume that they 
started in September, 1925, always making their quarterly 
deposit at least three days before the next interest day. 
How much money will be to their credit on Jan. 2,1935 ? 

Periodic Interest 

180. If interest which is due quarterly, or annually, is 
unpaid, some contracts provide that interest shall be paid 
upon unpaid interest as well as on the principal until all is 
paid. This form of interest is called Periodic Interest. This 
form of interest is illegal in some states. 

Example. What amount is due at the end of three 
years and six months on a debt for $ 500, with interest at 
6 °Jo payable semi-annually, on which at maturity no inter¬ 
est payments have been made ? 


268 


NEW HIGH SCHOOL ARITHMETIC 


The solution can be arranged in the following form: 



Interest 

Payments 

Due at 
End of 

Overdue 

Interest on 
Payment 

Amount of Pay¬ 
ment and Interest 

First 

$15 

6 mo. 

36 mo. 

$2.70 

$17 

70 

Second 

15 

12 mo. 

30 mo. 

2.25 

17 

25 

Third 

15 

18 mo. 

24 mo. 

1.80 

16 

80 

Fourth 

15 

24 mo. 

18 mo. 

1.35 

16 

35 

Fifth 

15 

30 mo. 

12 mo. 

.90 

15 

90 

Sixth 

15 

36 mo. 

6 mo. 

.45 

15 

45 

Seventh 

15 

42 mo. 

0 


15 

00 

Principal due at end of 42 mo. 

$500 

00 

Total amount due at end of 42 mo. 

$614 

45 


EXERCISE 119 

Assume periodic interest allowable in each of these ex¬ 
amples. 

1. What amount is due at the end of 2 yr. 6 mo. on a 
debt of $ 900, with 6 % interest, payable annually, on which 
no payments have been made ? 

2. What amount was due on Aug. 5, 1933, on a debt of 
$ 360 dated Feb. 5,1929, with 6 % interest, payable annually, 
on which no interest payments had been made ? 

3 . What is the amount due at the end of 5 years on a 
debt of $1200 with interest at 6 %, payable annually, on 
which the first two interest payments were made when due ? 

4 . What is the amount due at the end of 4 years on a debt 
of $ 2200 bearing 5 % interest, payable semi-annually, on 
which the first five interest payments were made when due ? 

5 . What amount was due Nov. 12, 1933, on a debt of 
$875 dated Aug. 12, 1929, with interest at 6 %, payable 
semi-annually, on which the first three interest payments 
were made when due ? 



















INTEREST 


269 


Installment Payment of Debts 

181. Installment payment of debts bearing interest are 
quite common. 

Example 1. A man purchased a lot on March 1, 1933, 
for $ 550, paying $ 50 ca£h and agreeing to pay $ 25 on the 
first of each month thereafter, together with the accrued 
interest on that payment at 6 % from the date of purchase 
until the date of payment. The payments may be recorded 
in the following manner. 


Payment 

Interest 

on 

Payment 

Payment 

Plus. 

Interest 

Balance 

Due 

Date 

Debt 

Due 

Time 

3/1/33 

$550 

$50 

— 

— 

$50 

$500 

4/1/33 

500 

25 

1 mo. 

$.12* 

25.13 

475 

5/1/33 

475 

25 

2 mo. 

.25 

25.25 

450 


a. Rule a sheet as above and complete the record until 
the debt is entirely paid. 

b. How much will the man pay all together ? 


Example 2. Assume that the agreement for the purchase 
of the lot in the preceding example had been that the man 
should pay $50 cash and, thereafter, $25 per month on 
the first of each month, with the interest that has accrued 
on the balance of the debt during the preceding month. 
Rule off and complete a schedule like the following, continu¬ 
ing it until the debt is fully paid. 


Date 

Debt 

Time 

Interest 

on 

Balance 

Payment 

Due 

Total 

Due 

New Balance 
op Debt 

3/1/33 

$550 

1 — 

— 

$50 

$50 

$500 

4/1/33 

500 

1 mo. 

$2.50 

25 

27.50 

475 

5/1/33 

475 

1 mo. 

2.37 

25 

27.38 

450 

6/1/33 

450 

1 mo. 

2.25 

25 

27.25 

425 


How much will the man pay all together ? 





































270 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 120 

Rule off and complete a schedule of payments for: 

1. A debt of $ 3000 at 6 % starting May 1, 1933, pay¬ 
ments of $ 100 to be made on the first of each month, to¬ 
gether with accrued interest frofh May 1 on each pay¬ 
ment to the date of payment. Payments are to be made 
until the debt is canceled. 

2. A debt of $ 1500 with interest at 6 °J 0 , dated Aug. 1, 
1933, payments of $ 50 to be made on the first of every 
month thereafter, until the debt is paid. On the first of 
each month, there is also to be paid the interest on the un¬ 
paid balance of the debt for the preceding month. 

3. A debt of $ 1325 for an automobile was to be paid as 
follows: The date of purchase was May 15, 1933; the 
initial payment was $125; the future payments were to 
be $ 100 on June 1, 1933, and a like amount on the first of 
each month thereafter until the debt was paid, together with 
interest at 6 % on each of said payments from May 15, 
1933, until the date on which the payment was made. 

4 . A street assessment of $ 650, dated Sept. 1, 1932, was 
to be paid as follows: on April 1, 1933, and on each suc¬ 
ceeding April 1 until the assessment and interest was fully 
paid, there was to be paid one tenth of the assessment, to¬ 
gether with accrued interest at 6 % on the unpaid balance of 
the assessment for the preceding year. 

5 . Assume a debt of $ 1000, dated March 15,1933, bearing 
interest at 6 %, to be paid as follows : a payment of $ 100 
to be made semi-annually; out of each payment enough was 
to be taken to pay the accrued interest for the preceding 
half year on the unpaid balance of the debt; the rest of the 
payment was to be used to decrease the balance of the debt. 

6 . A man bought a piano for $ 425, paying down $ 25, 
and agreeing to pay $ 10 per month. He also agreed to 


INTEREST 


271 


pay each, month the interest at 6 % that had accrued on the 
payment made from the date of purchase, May 20, 1933, to 
the date on which he made the payment. Assume that the 
payments were made on the 20th of each month. Find the 
total of all payments plus interest. 

7 . A man borrowed $ 4000 on Aug. 15, 1931, agreeing to 
pay interest at 6 % semi-annually. He was given the privi¬ 
lege of paying on the face of the debt on interest payment 
dates $ 100 or multiples thereof. On Feb. 15, 1932, he paid 
his interest and $ 300 on the debt; on Aug. 15, 1932, he 
paid his interest on the balance, and $200 on the debt; 
on Aug. 15, 1933, he paid the interest on the balance of the 
debt, and $ 600 to decrease the amount of the debt. Make 
out a schedule of interest and principal payments, assuming 
that the man plans to pay semi-annually the interest on the 
balance of the debt and also $ 250 toward the principal, after 
Aug. 15, 1933, until the debt is fully paid. Find the total 
of all payments made when the debt is fully paid. 


X. INSURANCE 

182. Property is often destroyed by fire, wind, and acci¬ 
dents ; individuals lose wages as a result of sickness, acci¬ 
dents, and other causes ; families lose their source of support 
through the death of some member. All persons are exposed 
to these risks, some more and others less, but only a small 
number actually suffer losses of any one kind in a limited 
period of time. 

Insurance is the business devised for distributing these 
losses so that each member of a group bears a share deter¬ 
mined by the protection he receives. The business of fur¬ 
nishing insurance is conducted by insurance companies who 
operate through agents to be found in all towns. 

Thus, 100 farmers each owning a $ 3000 barn might agree that 
they would rebuild the barn of any one of their number if it 
should be destroyed by fire. If one barn is destroyed, $ 3000 is 
lost. Each farmer would then contribute $ 30, including the man 
whose barn was lost. Each is willing to make this expenditure 
to protect himself against the possible loss of $ 3000. 

An Insurance Policy is the contract between the company 
and the insured. It states the amount of protection, called 
the Face of the Policy, and the conditions under which the 
insurance is afforded. Every one taking out an insurance 
policy should read it with great care, and should insist 
upon having the meaning of all its conditions fully ex¬ 
plained. 

An Insurance Premium is the sum which the insured pays 
the company in return for the insurance. 

272 


INSURANCE 


273 


Fire Insurance 

183. The Standard Fire Insurance Policy provides insur¬ 
ance “ to an amount not exceeding ” a specified sum for any 
direct loss or damage by fire. This loss is the “ actual cash 
value ” of the property damaged at the time of the fire. 

Note. — In case some insurance is paid an insured because of 
damage to his property by fire, the face of the policy is then auto¬ 
matically decreased by the amount of insurance already paid by 
the company. 

By agreement with the companies, an insured may divide 
his total insurance among two or more companies. Then, 
in case of loss, each company pays as much of the loss as 
its insurance is of the total insurance. 

Thus, if the total insurance is $ 5000, one company insuring 
$ 3000, and another $ 2000, then the first company would pay | 
and the second f of any loss by fire. 

184. The fire insurance premium depends upon many 
factors, — such as, the kind of fire protection provided by 
the city or town, the kind of building occupied by the 
insured, the business conducted by the insured, the number 
of occupants in the building, etc. As a rule, fire insurance 
companies are members of one of two associations of such 
companies. These associations employ men whose business 
it is to decide upon rates which will be adequate and just. 
These rates are recommended to the companies. Specific 
rates are recommended for each risk in a business district 
in a city, and a schedule of rates for all other kinds of 
insurable property. 

The rate is usually expressed as a number of cents per 
$ 100 of insurance for one year. 

185 Upon some kinds of property, a policy can be taken 
out for a term of three years or of five years. When a 


274 


NEW HIGH SCHOOL ARITHMETIC 


policy is taken out for a term of three years, the premium 
is two and one-half times , and for five years the premium is 
four times the annual premium. (Western Union rates.) 

Note. — Formerly the three-year premium was twice, and the 
five-year premium was three times the annual premium. 

Below is part of the dwelling house schedule used in a 
certain city. 

Rate per $ 100 of Insurance for One Year 


Dwelling House 

Brick — Shingle Roof 

Frame — Shingle Roof 

Number of 

Building 

Contents 

Building 

Contents 

Occupants 

alone 

alone 

alone 

alone 

One family . . . 

$.18 

$.22 

$.22 

$.26 

Two family . . . 

.20 

.24 

.24 

.28 

Three family . . . 

.22 

.26 

.28 

.32 

Barns and garages . 

.30 


.40 



186. The agent of the company receives a Commission 
for his services. 

The commission is a per cent of the premium charged by 
the company. 

One group of companies pays its agents: 

25 °!o on dwelling houses and contents and most public build¬ 
ings ; 20 °Jo on mercantile buildings; 15 % on other buildings. 


EXERCISE 121 

Determine the premium and the agent’s commission at 
25 % of the premium for each of the following policies, 
using the schedule above. 














INSURANCE 


275 


Ex. No. 

Face of 
Policy 

Term of 
Policy 

Property 

Insured 

Kind of 
Building 

Number of 
Families 

1 . 

$2750 

1 yr. 

building 

brick 

two 

2 . 

3200 

3 yr. 

building 

frame 

three 

3 . 

1200 

5 yr. 

contents 

frame 

one 

4 . 

500 

3 yr. 

contents 

brick 

two 

5 . 

4500 

5 yr. 

building 

frame 

one 

6 . 

1500 

5 yr. 

contents 

frame 

three 

7 . 

750 

3 yr. 

garage 

frame 


8 . 

•650 

5 yr. 

contents 

frame 

three 

9 . 

1500 

5 yr. 

garage 

brick 


10 . 

3750 

3 yr. 

building 

frame 

one 


11 . A factory valued at $ 25,000 is insured for 80 % of 
its value at an annual rate of 70^ per $ 100. 

a. What is the premium ? 

b. What is the agent’s commission if he receives 15 % 
of the premium ? 

c. How much does the company actually get ? 

12. A store owner asks an agent to write up for him 
$30,000 worth of insurance on his store, —the insurance 
to be in three companies. 

The agent gives company A $ 12,000, company B $10,000, 
and company C, $ 8000. The rate in each company is 55 ^ 
per $ 100. Determine : 

a. The premium the owner must pay; 

b. The commission earned by the agent at 20 % of the 
premium; 

c. The amount each company receives. 

13. A factory valued at $ 25,000 is insured for 80 % of 
its value at an annual rate of 70 ^ on $ 100. 

a. Determine the premium for three years. 

b. What is the agent’s commission if he receives 15 % 
of the premium ? 

c. What does the company receive ? 













276 


NEW HIGH SCHOOL ARITHMETIC 


14 . Assume that fire causes actual cash loss of $ 5000 to 
the property insured in Ex. 12. How much must each of 
the companies pay the insured ? 

15 . A corporation insures its $ 30,000 plant for 75% of its 
value at 45 4 per $ 100 in three companies. Company A takes 
$ 10,000, company B $ 5,000, and company C, the balance. 

a. What is the face of the policy with company C ? 

b. What premium is paid each company ? 

c. The companies each allow the agent 15 %. What are 
his commissions ? 

d. How much does each company actually receive ? 

e. A fire causes a $ 10,000 loss. How much must each 
company pay the corporation ? 

16 . The owner of a house valued at $ 7500 insured it 
for three years for $ 5000 at the rate of 25 4 per $ 100 for 
one year. 

a. What premium did he pay ? 

b. The agent’s commission was 25 %. What did he 
receive ? 

c. The house was completely destroyed by fire. How 
much should the owner receive from the company ? 

17 . A man insured his house for $ 8000 at the annual 
rate of 28 4 per $ 100, and his household goods for $ 2500 
at 36 4 per $ 100. 

a. Determine his total premium. 

b. His house is completely destroyed by fire, but $ 850 
worth of the household goods are saved. How much insur¬ 
ance did he receive ? 

18 . Bind the premium for a $ 1000 policy on a farmer’s 
barn at the rate of 60 4 per $ 100 per year, the policy to 
run for 5 years. 

19 . If a farm is occupied by a tenant, the annual rate on a 
farm policy is 15 4 higher than that in Ex. 18. Find the pre¬ 
mium for the policy in Ex. 18, if the occupant is a tenant. 


INSURANCE 


277 


187. Short Term Rates. —Policies are sometimes written 
for less than a year. In such cases the following table of 
short term rates is used. 

These rates are used also when a policy is canceled. If 
the insured asks to have the policy canceled, and he always 
has the right to do so, the company is allowed to charge 
the short term rate for that part of the term of insurance 
which has already expired, and must return to the insured 
the balance of the original premium. If the insurance com¬ 
pany cancels the policy, and it has the right to do so on 
five days’ notice, then it is allowed to charge for that part 
of the term of insurance which has already expired as much 
of the original premium as the expired term is of the orig¬ 
inal term. 

Thus, if a one-year policy has run for seven months, the com¬ 
pany can retain only seven-twelfths of the original premium. 
This is called pro-rating the premium for the short term. 


Short Rate Table for One-Year Policies 


Time 
in Days 

Charge 

Time 
in Days 

Charge 

Time 
in Days 

Charge 

Time 
in Days 

Charge 

1 

2% 

14 

13% 

55 

29 

180 

70% 

2 

4 % 

15 

13% 

60 

30% 

195 

73% 

3 

5% 

16 

14% 

65 

33% 

210 

75% 

4 

6% 

17 

15% 

70 

36% 

225 

78% 

5 

7% 

18 

16% 

75 

37% 

240 

80% 

6 

8 % 

19 

16% 

80 

38% 

255 

83% 

7 

9 % 

20 

17% 

85 

39 % 

270 

85% 

8 

9 % 

25 

19% 

90 

40% 

285 

88% 

9 

10 % 

30 

20% 

105 

46% 

300 

90% 

10 

10% 

35 

23% 

120 

50% 

315 

93% 

11 

11% 

40 

25% 

135 

56% 

330 

95% 

12 

11% 

45 

27% 

150 

60% 

345 

98% 

13 

12% 

50 

28% 

165 

66 % 

360 

100% 




















278 


NEW HIGH SCHOOL ARITHMETIC 


Example. If the annual rate is 65 then the short rate 
for 85 days is 39 % of 65 or 25.35 

If the number of days does not appear in the table, take 
the per cent opposite the next greater number of days that 
does appear in the table. 

Thus, the per cent for 87 days is that opposite 90 days, namely 
40 <Jo. 

A similar table of short rates is provided for term 
policies. 

EXERCISE 122 


Using the table of short term rates, find the premium : 


Ex. 

No. 

If the 
Face Is 

If tjie 
Rate Is 

If the Term Is 

a 

6 

c 

d 

1 . 

$3500 

$.40 

1 mo. 

3 mo. 

7 mo. 

9 mo. 

2 . 

7200 

.55 

20 da. 

2 mo. 

5 mo. 

75 da- 

3 . 

850 

.75 

15 da. 

4 mo. 

35 da. 

225 da- 

4 . 

1300 

.90 

25 da. 

80 da. 

200 da. 

11 mo. 

5 . 

9500 

.65 

10 da. 

92 da. 

165 da. 

260 da. 


Determine the original premium and the refund to the 
insured in each of the following cases : 


Ex. 

No. 

Face of 
Policy 

Term 

Rate fob 
the Term 

Canceled 
at End of 

By Whom 
Canceled 

6 . 

$1200 

1 yr. 

$.60 

4 mo. 

Company 

7 . 

750 

1 yr. 

.88 

7 mo. 

Company 

8 . 

2000 

1 yr. 

1.12 

160 da. 

Company 

9 . 

3200 

1 yr. 

.45 

210 da. 

Insured 

10 . 

1350 

1 yr. 

.65 

185 da. 

Insured 

11 . 

650 

1 yr. 

.72 

96 da. 

Insured 

12 . 

4200 

1 yr. 

.68 

163 da. 

Insured 

13 . 

5500 

1 yr. 

.88 

155 da. 

Company 

14 . 

2250 

1 yr. 

.95 

225 da. 

Company 































INSURANCE 


279 


15 . A man who has mortgaged his house and lot as 
security for a loan of $ 3000 for 6 months agrees with the 
lender to insure the house for $ 3000 for the six months. 
The annual rate is $ .60. Find the premium for six 
months. 

16 . The owner of a store insured the building and con¬ 
tents for $ 12,500 at $ .75 per $ 100 for one year. 

a. What was the premium ? 

b. At the end of three months the company canceled the 
policy. How much were they required to refund to the 
insured ? 

c. Suppose that the insured had asked to have the policy 
canceled, what would have been the refund ? 

17 . A man insured his household goods for $ 2500 for 
three years at the rate of $ .22 per $ 100 per year. 

a. What was the premium ? 

b. How much did the agent receive if his commission 
was 15 % ? 

c. How much did the company receive ? 

d. At the end of fifteen months, he asks to have the 
policy canceled. If the short rate for fifteen months of a 
three-year term policy is 48 % of the whole premium, how 
much must the company refund to him ? 

18 . A man has his property insured for $ 1800 for three 
years at the rate of $ .28 per $ 100 per year. 

a. What was his premium ? 

b. At the end of one year, he wants the face of the policy 
changed to $ 2500 and is told by the agent that there is 
now a new rate of $ .32 per year. The agent agrees to 
cancel the old and write the new policy, prorating the 
premiums. How much additional must the insured pay 
for the balance of the three-year term ? 


280 


NEW HIGH SCHOOL ARITHMETIC 


19. A family insure their household goods for three 
years for $ 2000 at the annual rate of $ .22. At the end 
of one year, they move into a house where the annual rate 
is $ .26. How much additional premium must they pay if 
the company prorates each of the premiums ? 

20. On Sept. 11, 1931, a man insured his household 
goods for one year for $ 1500 at the rate of $ .44. 

a . What was his premium ? 

b. On Feb. 26, 1932, the man moved into a house where 
the rate was $ .38 per year. How much refund should he 
have received from the company ? (Prorate premiums.) 

188. Co-insurance. — Some policies provide that the in¬ 
sured should insure his property for a specified per cent of 
the value of the property. The owner, however, may not 
take out the specified amount of insurance. In such cases, 
the policy provides that only as much of any loss suffered 
will be paid as the actual insurance is of the specified in¬ 
surance, not exceeding, however, the face of the policy. 

Thus, suppose that a piece of property should be insured for 
$ 8000 and that the owner carries only $ 6000 worth. Then the 
owner will receive only f of any loss he suffers by fire. If the 
fire damage is $4000, he will receive $3000. 

This provision is contained in a co-insurance clause of the 
policy, which states the per cent of insurance that the in¬ 
sured should carry. The idea is that the insured becomes a 
co-insurer with the company if he does not take out the 
specified amount of insurance. Co-insurance is never ap¬ 
plied to dwellings or household goods. Usually a lower 
rate is charged when the policy contains a co-insurance 
clause. 

EXERCISE 123 

1. How much insurance will a man receive whose 
$ 20,000 property is insured for $ 10,000 under a policy 


INSURANCE 


281 


containing an 80 °J 0 co-insurance clause when the loss by 
fire is : a. $ 1200 ? b. $ 2500 ? c. $ 6500 ? 

2. How much insurance should a man receive whose 
$ 8000 property is insured for $ 4000 under a policy con¬ 
taining a 75 % co-insurance clause if the damage by fire is: 
a. $2100? b. $1200? c. $7000? 

3. How much would the owner in Example 2 receive 
under a standard policy in each of these cases ? 

4 . A store, valued at $ 6000, is insured for $ 3500 at 
$ .65 per year, and the contents, valued at $ 15,000, is in¬ 
sured for $ 7500 at $ .80, both in companies whose policies 
contain 80 % co-insurance clauses. The term of the policy 
is three years. A fire causes $ 500 damage to the store 
and $ 2000 damage to the contents. 

a. What was the total premium ? 

b. How much insurance is paid the owner on the store 
and on the goods ? 

5. Property valued at $ 1000 is insured for $ 600 with 
a 90 % co-insurance clause. A fire causes a loss of $ 875. 
How much insurance does the owner receive from the 
company ? 


EXERCISE 124 
Supplementary Exercises 

Before doing the following examples, obtain from some 
fire insurance agent in your own community the rates 
needed in solving the examples. Also learn whether the 
policy is a standard policy or one containing a co-insurance 
clause. Then find the premium for : 

1. A $ 5000 policy for three years on a two-story brick 
building valued at $ 7200 if there are adjacent to it other 
brick buildings, if the first floor is occupied by a grocer 
and the second floor by doctors’ offices* 


282 


NEW HIGH SCHOOL ARITHMETIC 


2. A policy for $ 15,000 for five years on a frame church 
valued at $ 17,500. Assume that there are not any other 
buildings within fifty feet of it and that the church is 
heated by a furnace. 

3. A policy for $ 25,000 for three years on a two-story 
brick school building valued at $ 40,000, heated by steam. 

4 . A policy for $ 5000 for three years on a two-story 
frame dwelling having a stucco exterior, valued at $ 7250. 

5 . Suppose that the church is damaged to the extent of 
$ 800. How much insurance must the company pay ? 

Other Property Insurance 

189. Property is insured against loss by wind, called 
Tornado Insurance ; against loss at sea, called Marine Insur¬ 
ance; against theft, called Burglary Insurance, etc. The 
following examples are some of the simple ones that occur. 
The rates are in every case bona fide, having been taken 
from recent agents’ manuals. 


Automobile Insurance 

The rates and conditions for automobile insurance of all 
kinds vary so much in different parts of the country, and 
are changed so frequently, that it is unsatisfactory to quote 
such rates in a text, except for illustration. 

There are three kinds of automobile insurance. 

a. Fire and theft insurance , as protection to an owner if 
his car should be stolen or damaged by fire. 

h. Property and public liability insurance , to protect an 
owner in case of claims made against him for damages to 
property or to other persons. 

c. Collision insurance , to protect an owner for some or all 
of the damage done to his own car by some other person. 


INSURANCE 


283 


The following premiums were charged in a recent year in 
one community for automobile fire and theft insurance: 


Age of Cab 

Ford 

Buick Eight 

Packard Eight 

New car . . . 

$4.70 

$7.40 

$11.35 

1 yr. old . . . 

3.55 

5.60 

8.55 


For $ 5000 of property damage insurance, the premium 
was $ 8 per year. For $ 5000 public liability insurance on 
each of one or two persons, the premium was $ 17. 


EXERCISE 125 

Find the total premium for fire, theft, property damage, 
and liability insurance on: 


Ex. No. 

The Car 

Age 

Ex. No. 

The Car 

1 . 

Ford 

New 

4 . 

1 yr. old Ford 

2 . 

Buick Eight 

New 

5 . 

1 yr. old Buick 

3 . 

Packard Eight 

New 

6 . 

1 yr. old Packard 


7. The owner of a bookstore takes out $ 3000 of burglary 
insurance at the rate of 2f %. Find the premium. 

8 . The owner of an auto supply store wants $ 4000 worth 
of burglary insurance. He is told that the annual rate is 
6.6 % with a discount of 5 % if there is a watchman in the 
building. Find his rate if he can claim the discount. 

9. A bank is offered the rate of 3 % for burglary insur¬ 
ance with a discount of 15 % because of the size of the town, 
and 15 % additional because it has a watchman. Find the 
annual premium for $ 15,000 worth of insurance. (The two 
discounts are treated as successive discounts as in § 155.) 

10. A house is insured against damage by wind for $ 1500 
at the rate of f % for three years. Find the premium. 




















284 


NEW HIGH SCHOOL ARITHMETIC 


11 . A farmer’s bam is insured for $ 2500 against loss or 
damage by fire, lightning, or tornado, at the rate of 2 \°/o> 
Find the premium. 

12. The owners of some greenhouses take out insurance 

against loss by wind, or hail, for $ 3500 for three years 
with a rate of f % for the whole term for the tornado in¬ 
surance alone and additional for the hail insurance. 

Find the premium. 

13 . A farmer insures his three horses for $ 150 each and 
his five cows for $ 75 each for one year, against injury or 
loss due to fire, wind, or lightning, at the rate of 1^%. 
Find the premium. 

14 . A farmer insures his frame 'silo against loss due to 
wind for $1750 for five years at 6 % for the term. Find 
the premium. 

15 . A large ranch owner insures his new tractor for 
$ 1500 against damage by fire for 1 % for one year, and 
against damage by tornado for \ % for one year. What 
is the combined premium ? 

Life Insurance 

190. Life Insurance premiums are based upon very elabo¬ 
rate computations in which percentage in one form or 
another plays an important part. These computations, 
however, are all made in the actuarial departments of the 
companies. The computations which either the agent or 
the insured have to make are quite elementary. The pre¬ 
miums depend upon the kind of policy the insured desires 
and upon the age of the insured. * Below are given the 
rates of a certain company for certain ages for each of 
three kinds of policies. Similar rates for applicants of all 
ages between 20 and 60 inclusive are obtainable from the 
agents of a company. The rates of different companies 
vary a little but not much from these. 


INSURANCE 


285 


Life Insurance Rates 


Age . 

Ordinary Life 
Policy 

20-Payment Life 
Policy 

20-Year Endowment 
Policy 

20 . 

$ 18.69 

$27.66 

$48.93 

21 . 

19.09 

28.08 

49.00 

22 . 

19.52 

28.54 

49.08 

23 . 

19.95 

29.02 

49.18 

24 . 

20 40 

29.51 

49.28 

25 . 

20.80 

30.02 

49.37 

26 . 

21.41 

30.55 

49.48 

27 . 

21.93 

31.10 

49.60 

28 . 

22.50 

31.66 

49.72 

29 . 

23.09 

32.26 

49.85 

30 . 

23.72 

32.87 

49.98 


In the first column is the nearest age of the applicant. 
Opposite the age are the rates for the three different 
policies. 

Under the ordinary life policy, the insured pays premiums as 
long as the policy is in force; the face of the policy is payable to 
the beneficiary at the death of the insured. 

Under the twenty-payment life policy, the insured pays pre¬ 
miums for twenty years at most. At the end of twenty years, the 
policy is fully paid. The face of the policy is payable to the 
beneficiary of the insured at the death of the latter, whether that 
occurs before or after twenty years. 

Under the twenty-year endowment policy, the insured pays 
premiums for twenty years at most. The face of the policy is 
payable to the insured at the end of twenty years, if he is living, 
or to the beneficiary of the insured if the latter should die before 
twenty years. 

If the premium of a life insurance policy is paid semi¬ 
annually, the semi-annual premium is found by adding to 
the annual premium six per cent of itself and dividing the 
result by two. 













286 


NEW HIGH SCHOOL ARITHMETIC 


If the premium is paid quarterly, the quarterly premium 
is found by adding to the annual premium four per cent of 
itself and dividing the result by four. 

EXERCISE 126 


What is the annual premium for: 


Ex. No. 

Kind op Policy 

Amount 

At Age 

1 . 

Ordinary life 

$ 3000 

28 

2. 

20-payment life 

5000 

27 

3. 

20-year endowment 

2500 

25 

4. 

20-payment life 

4500 

30 

5. 

Ordinary life 

10,000 

29 

6. 

20-year endowment 

3500 

22 

7. 

20-payment life 

2500 

26 

8. 

Ordinary life 

1500 

23 

9. 

20-year endowment 

4500 

30 

10. 

20-payment life 

7500 

29 


11-20. Determine the semi-annual premiums for each of 
the policies. 

21-30. Determine the quarterly premiums for each of 
the policies. 

191. Dividends. —Most life insurance policies now issued 
are participating policies. Annually the insured receives 
from the company what is incorrectly called a dividend. 
Actually, it is a refund from the company of the insured’s 
share of savings made by the company on estimated ex¬ 
penses, or of savings from estimated losses, or of certain 
interest earnings of the company in excess of those antici¬ 
pated. These dividends may be used by the insured to 
decrease the amount of the premium due the company, may 
be allowed to accumulate at compound interest in the care 
of the company, or may be disposed of in other ways with 
the consent of the insured and of the company. 











XI. TAXES 


192, Village, city, town, county, and state taxes include 
some or all of the following: 

a. Real Estate Tax, levied upon the assessed value of 
the real estate (including improvements) of an individual. 

b. Personal Property Tax, levied upon the removable 
personal property, such as household goods, automobile, 
animals, money, securities, etc. 

c. Special Taxes for special purposes, such as the con¬ 
struction of sidewalks, sewers, streets, etc. 

d. Income Tax, levied upon the assessable income of an 
individual. 

e. Inheritance Tax, levied upon wealth inherited from a 
deceased resident of the taxing community. 

/. Poll Tax, a fixed amount levied upon each (male) 
resident of a community. 

Of these taxes, the first three are quite general; the next 
two are increasingly becoming sources of revenue for gov¬ 
ernmental agencies. 


Property Taxes 

193. The assessed value of property is the valuation 
placed upon it for taxing purposes by the properly con¬ 
stituted officers, called Assessors. 

The value is sometimes definitely assumed to be only 
one third or one half the actual value; in other cases, it is 
assumed to be the actual value of the property. 

Any individual, dissatisfied with the assessed valuation 
of his property, may appeal to the Board of Review, at an 
agreed-upon time, for a revaluation. 

287 


288 


NEW HIGH SCHOOL ARITHMETIC 


194. The tax itself is computed as a certain rate per 
cent of the assessed value; this rate is called the Tax Rate. 

The following example shows how one community deter¬ 
mined its tax rate in a recent year. 

Example 1. 

The Assessable Property 


Assessed value of real property.$ 57,791,875 

Assessed value of personal property. 12,911,365 

Total assessed value of property.$ 70,703,240 

Total Taxes Required 

The city’s share of state taxes.$ 87,952.82 

The city’s share of general county taxes. 99,862.89 

City taxes for educational purposes. 281,987.05 

City taxes for indebtedness. 172,035.00 

City taxes for street improvements. 85,761.94 

City taxes for general purposes. 262,245.66 

Total taxes required.$ 989,845.36 


The tax rate was determined by regarding the total 
taxes, $ 989,845.36, as the percentage and the total assess¬ 
able property, $70,703,240, as the base. 

What per cent of $ 70,703,240 is $989,845.36? The re¬ 
sult is the tax rate. 

By § 151, the tax rate is 1.4%. 

Example 2. What was the real estate tax and the 
personal property tax of a man whose real estate was 
valued at $ 3575 and personal property at $ 1450 ? 

Solution. 1. The realty tax = 1.4 °/o of $ 3575 

= $ 50.05 

2. The personal property tax = 1.4 °Jo of $ 1450 

= $ 20.30. 

Where the assessed valuation of the property is very 
great, as in the case of railroads and public utility corpora- 














TAXES 


289 


tions, it is necessary to carry ont the tax rate to several 
decimal places. 

Thus, in a recent year, one state published as its rate for state 
tax purposes 1.317260878 %. 


EXERCISE 127 

Determine the total property taxes of the following 
persons: 


Ex. No. 

Name of Person 

Real 

Property 

Personal 

Property 

Where the Tax 
Rate Is 

1 . 

Henry Radley 

$ 1525 

$ 780 

1.5% 

2. 

Martin Fosdick 

— 

550 

1.8% 

3. 

Samuel Girard 

4650 

20,500 

1.3% 

4. 

Fred Thompson 

7275 

6,825 

1.7% 

5. 

William Peterson 

19,725 

32,650 

1.4% 

6. 

Frank O’Malley 

6230 

1245 

11 mills on $1 

7. 

James Sargent 

19,750 

3235 

16 mills on $1 

8. 

Ajax Stone Co. 

12,450 

2850 

13 mills on $1 

9. 

Edwards Grocery Co. 

8750 

7450 

19 mills on $1 

10. 

Peter Schmidt 

2385 

675 

18 mills on $1 


11. A certain town whose taxable property is $ 1,150,000 
wishes to raise $ 15,100 by taxation. What will be the 
rate of taxation expressed in mills? What will be the 
entire tax of an individual whose property is valued at 
$ 8500 ? 

12. If a town whose taxable property is $ 75,400 wishes 
to raise $ 1055.60, what is the rate of taxation ? Express 
the result as a per cent, carried out to one decimal place as 
in Examples 1-5. 

13. A town whose taxable property is $ 165,000 wishes 
to raise $ 3630. What is the rate of taxation ? 

14. A man’s house is valued at $ 3640. What will be 
his tax, at the rate of $ 14.50 on $ 1000 ? 













290 


NEW HIGH SCHOOL ARITHMETIC 


15. a. A town whose real estate is valued at $ 724,600 
and personal property at $ 561,400 wishes to raise $ 16,000 
in taxes. What must be its tax rate ? (Carry out to two 
decimal places as in Ex. 15.) 

b. Eind the taxes of a man whose total property is 
valued at $ 7365. 

16. In one community the assessed valuation of property 
was about \ the real value and the tax rate was 5.8 % ; in 
another, the assessed valuation was about 90 % of the real 
value, and the tax rate was 1.4 %. Compare the taxes of 
two men, one living in each community, each of whom had 
property of actual value $ 8000. 

17. a. In a certain state, the assessed valuation of rail¬ 
road property in 1916 was $ 360,960,000 and the tax rate 
was 1.3172+ %. 

Determine the total taxes contributed by the railroads. 

b. The valuation of all general property in the state was 
$ 3,426,797,220. 

Determine the aggregate of all the property taxes of the 
state if the average rate of taxation was 1.3172'*' %. 

18. In this state 85 % of the taxes collected by the state 
from street railways is returned to the community in 
which the street railway is located, and 15 % is retained by 
the state. 

One street railway was assessed at $ 850,000. Determine 
its total tax (rate= 1.3172+%), the amount retained by the 
state, and the amount returned to the community. 

19. A man bought a house and lot for $ 5500. It was 
assessed for 85 % of its value and taxed at the rate of 16 
mills on $ 1. His insurance and repairs amounted to $ 50 
per year. He rented the house for $ 40 per month. What 
per cent of income on his investment did he receive ? 

20. A man owns a two-family building that cost him 
S 7350, on a lot which cost $ 1150. His property is assessed 


TAXES 


291 


at about 60 % of its value and taxed at tbe rate of 171- 
mills on $ 1. He estimates insurance and repairs at $ 100 
per year. What total rent must lie receive per month to 
make 8 % annually on his investment ? 

21. A man paid $5200 for a house and lot. It was 
valued by the assessor at $ 3850. The tax rate was 1.4 %. 
He estimates his annual charge for insurance at $ 5.30, and 
for repairs and painting, $* 40. 

a. What was his annual “ rent/ 7 including interest at 6% 
on his investment, taxes, insurance, and repairs ? 

b. What was his monthly rent ? 

c. Before purchasing the house, he rented it at $ 40 per 
month. Did he gain or lose annually by purchasing the 
house ? How much ? 

d. After owning it four years, he sold the property for 
$ 5800. What percentage of annual profit on the $ 5200 
investment do his profits under parts c and d represent ? 

e. Since he has included in part a interest at 6 % as 
part of his rent, what is his total rate of income per year 
on his $ 5200 investment ? 

22. What tax will be paid by a man whose real estate is 
valued at $ 5200, and whose personal property is valued at 
$ 3850 if he also pays for two polls at $ 1.50 each, the rate 
of taxation being $ 16.50 on $ 1000. 

23. The assessed valuation of real estate in a certain 
town is $ 825,000 and of personal property $ 394,625. 
There are 465 citizens who pay a poll tax of $ 1.50 each. 
The town wishes to raise $ 16,475 in taxes. What must be 
the rate of taxation ? 

195. Special Assessments are levied at times upon the 
owners of property which will be benefited by some special 
improvement. They are levied particularly in villages and 
cities to obtain income with which to pay for the construc¬ 
tion of sidewalks, streets, sewers, etc. The computation 


292 


NEW HIGH SCHOOL ARITHMETIC 


of the assessments for street construction is usually based 
upon a specified amount per front foot. These and the 
other special assessments rarely present any arithmetical 
difficulties. 

Supplementary Topics 

196. Several states in the United States now derive rev¬ 
enue from Income Taxes. While the laws of the several 
states vary in minor particulars, all agree in specifying that 
a certain part of the income of a resident of the state shall 
be exempt from taxation, and secondly that income derived 
from certain sources shall be exempt, — such as income from 
corporations that pay taxes to the state. The balance of 
the income of a resident is called his taxable income. 

In one state, $ 800 of annual income of every resident 
is exempt, and $ 1200 of income of every married man 
whose wife is living; besides, an additional $ 200 for every 
child under 18 years of age, and for every dependent of the 
resident, is exempt. The rates of taxation for taxable in¬ 
come follow: 

Taxable income between $ 1 and $ 1000 is taxed 1 % ; 
between $ 1000 and $ 2000, 1\ % ; between $ 2000 and 
$ 3000, 11 % ; between $ 3000 and $ 4000, If % ; between 
$ 4000 and $ 5000, 2 % ; between $ 5000 and $ 6000, 2\ % ; 
and each additional $ 1000 is taxed an additional 1 °] 0 until 
$ 13,000 is reached. All income above $ 13,000 is taxed 6%. 

Note. — Observe carefully that a certain rate applies to the 
first $ 1000 ; then a new rate to the second $ 1000 ; a new one to 
the third $ 1000; etc., until $ 13,000 is reached. 

Example. What is the annual income tax of a man in 
this state whose taxable income is $ 2500 ? 

Solution. 

1. The tax on the first $ 1000 is 1 °Jo of $ 1000 or $ 10. 

2. The tax on the second $ 1000 is \\ °fo of $ 1000, or $ 12.50 

3. The tax on $ 500 in excess of $ 2000 is 1^ °Jo of 8 500 or $ 7.50 

4. .*. the total income tax is $ 30.00 



TAXES 


293 


EXERCISE 128 

1. Make out a table giving the annual tax on a taxable 
income of $1000, $2000, $3000, $ 4000, etc., to $13,000 
inclusive. 

2. Henry Smith, unmarried and without dependents, 
earns $ 95 per month of taxable income. What is his 
annual income tax ? 

3. a. Charles Williams, unmarried, but supporting his 
mother and ten-year-old brother, earned $ 95 per month 
before the war. What, if any, income tax did he pay an¬ 
nually ? 

b. Ten years later he was earning $ 165 per month. What 
income tax did he then pay ? 

4. Mr. Hughes, supporting his wife and two children, 
had an annual income from all sources of $ 7350. Of this, 
$ 675 was exempt, since it was derived from business outside 
the state. What was his income tax ? 

5. Edward Evans and his wife, jointly, have a total in¬ 
come of $ 23,750. They do not have any children under 
18 years of age. What is their annual income tax ? 

197. Corporations in that same state also pay an income 
tax according to the following general provisions. 

On the first $ 1000 of taxable income, 2 % ; on the second 
$ 1000 or part thereof, 2-j- % : on each succeeding $ 1000 or 
part thereof an additional \ tfc, until $ 7000 is reached. On 
all taxable income above $ 7000, 6 %. 

EXERCISE 129 

What is the income tax of a corporation in this state 
whose taxable income is : 

1. $2575 4. $5765 7. $142,650 10. $432,500 

2 . $3500 5 . $16,000 8 . $325,000 11 . $522,850 

3. $4200 6. $35,275 9. $ 175,000 12. $738,750 


294 


NEW HIGH SCHOOL ARITHMETIC 


State Inheritance Tax 

198. Some states derive income from inheritance taxes. 
The laws regulating these taxes also vary hut have some 
common elements. The rate of taxation depends upon the 
amount of the inheritance, and secondly, upon the relation-- 
ship of the beneficiary to the deceased. The following 
table gives the rates in one state, for near relatives. 


Beneficiary 

Amount 

Exempt 

Bate of Taxation on : 

First 
$ 25,000 

$ 25,000 
to 

$ 50,000 

$ 50,000 
to 

$ 100,000 

$ 100,000 
to 

$ 500,000 

Amounts 

over 

$500,000 

Husband or wife 

$10,000 

1 <fo 

1 l°?o 

2 °]o 

2 h</° 

3 °!o 

Children (each) 

$ 2,000 

1 % 

1 \°(o 

2 °to 


3 °/o 

Brother, sister, 

$500 

1 \°Jo 

2 \°Io 

3 °Jo 

3i7» 

4 \°!o 

nephew, niece 







Uncle, aunt, cousin 

$250 

3% 

4 

6 °fo 


9 °fo 


The tax is collected by the county in which the deceased 
lived. Out of the tax, T-J- % is retained by the county, and 
the balance is turned in to the state treasury. 

EXERCISE 130 

1. Out of an estate valued at $ 85,000, the following be¬ 
quests were made: to wife, $ 40,000; to daughter and son, 
each $ 20,000; to brother and niece each $ 2000; to aunt, 
$ 1000 . 

a. How much tax was levied against each beneficiary ? 

b. What amount was retained by the county and what 
amount was turned in to the state treasury ? 

2. Assume that the following bequests are made to near 
relatives out of an estate: 

To wife, $ 125,000 ; to son, $ 75,000; to daughter, $ 75,000; 
















TAXES 295 

to sister, $ 10,000 ; to nephew, $ 5000; to niece, $ 7500 ; to 
uncle, $ 3500 ; to aunt, $ 5000. 

a. What tax was levied against each beneficiary ? 

b. What amount was retained by the county ? 

National Taxes 

199. The money for the support of the national govern¬ 
ment is derived chiefly from Duties levied upon the most 
recent Tariff Law and taxes levied upon the most recent 

Revenue Act of Congress. 


Duties 

200. Duties or Customs are taxes levied upon goods im¬ 
ported into the United States. 

A Specific Duty is a fixed tax upon an article without 
regard to its value. If the weight or quantity of the goods 
is expressed in metric units, these must be changed to equal 
English units. 

Thus, the duty on maple sugar is 3 f per pound. 

An Ad Valorem Duty is levied at a certain rate per cent 
of the cost of the goods in the country from which they 
are imported. If this cost is expressed in units of foreign 
money, the cost must be expressed in units of our money. 

Thus, the rate on silk clothing is 50 °fo ad valorem. 

If a silk dress costs 500 fr. in France, the equivalent price in our 
money is 500 x $ .193 or $ 96.50. The duty on it would be 50 °fo 
of $96.50 or $48.25. 

Ad valorem duties are not computed on fractions of a 
dollar. 

201. When the duty is specific, certain allowances are 
made. 

Tare is an allowance made for the weight of the container 
in which the goods is imported. 


296 


NEW HIGH SCHOOL ARITHMETIC 


Leakage is an allowance for the loss of any liquids im> 
ported in casks or barrels. 

Breakage is an allowance for the loss of any liquid im¬ 
ported in bottles. 

202. The Tariff Law provides rates of taxation on all 
articles likely to be imported. 

Note. — A copy of the latest tariff law can be obtained through 
the representative in Congress from your district. 

There is a list of articles on which duties are levied, — 
called the Dutiable List; there is also a list of articles on 
which duties are not levied, — called the Free List. 

Thus, pure bred animals for breeding purposes, leather boots 
and shoes, coffee, enter free of duty. 

203. Administration of the Tariff Law. — Goods purchased 
abroad must enter the United States through certain ports 
called Ports of Entry. At each such port, is located a 
United States Customs House. Proper officers at the cus¬ 
toms house receive a description of important merchandise, 
examine some or all of a shipment, decide upon the duties 
that must be paid, and, after receiving the duties, issue the 
necessary permits to have the merchandise landed. 

Note. — Importers may store goods in government warehouses, 
after giving bond for the payment of the duties. Later, upon 
payment of storage charges, duties, etc., the importer may with¬ 
draw his goods from the warehouse. 

Individuals entering the United States from abroad must 
declare whether they have in their baggage any dutiable 
goods. Their baggage and persons are subject to inspection. 

EXERCISE 131 

What is the duty on: 

l. 20 doz. watches @ 1275 fr. per doz v duty 30 % ad 
valorem ? 


TAXES 297 

2. 1 gr. gold rings @ 14 sh. for each ring, duty 60 % ad 
valorem ? 

3. 1 fur cape valued at £ 35, duty 40 % ad valorem ? 

4. An invoice of linen goods valued at £ 462, duty 40 % 
ad valorem ? 

5. 150 meters silk dress goods @ 15 fr. per meter, duty 
45 % ad valorem ? 

6. 100 doz. table knives @ £ 1 per doz., duty 30 % ad 
valorem ? 

7. 300 gal. wine @ 20 fr. per gallon, duty 45 % ad 
valorem ? 

8. 500 T. sugar, duty <£ per pound ? 

9. On 65 small oriental rugs having an average value of 
1000 fr. each, duty 45 % ad valorem ? 

10. On 15 silk handkerchiefs valued at 3 sh. each, duty 
45 % ad valorem ? 

11. On a diamond ring valued at £ 53, duty 60 % ad 
valorem ? 

12. On 100 meters lace @ 3 fr. per meter, duty 60 % ad 
valorem ? 

13. On 5 doz. pairs hose @ £ 1 per dozen, duty 40 % ad 
valorem ? 

14. On 150 pieces plate glass 20" X 22", 100 pieces 
24" X 28", and 100 pieces 24" x 32", duty 8/ per square 
foot ? 

15. On 750 lb. camphor, duty 1^ per pound; 50 lb. 
benzoate of soda, duty 5^ per pound; and 75 lb. menthol, 
duty 50 / per pound ? 

16. On a telescope valued at 375 fr., duty 25 % ? 

17. On a marble statuette valued at £ 69, duty 45 % ? 

18. On an automobile valued at 12,500 fr., duty 45 % ? 

19. On 15 lb. of tobacco, duty & 2.50 per pound ? 

20. On a suit of clothes valued at £ 5 8 sh., duty 35 % ! 


298 


NEW HIGH SCHOOL ARITHMETIC 


Internal Revenue 

204. Revenue taxes for the Government derived from 
sources within the country are levied at present (1933) 
under a Revenue Act of Congress which was enacted in 
June, 1932. This act levies taxes upon the manufacture, 
sale, and consumption of various products; upon various 
forms of amusement and business; upon incomes of indi¬ 
viduals and corporations ; and upon inheritances. 

Note. — A copy of this Act of Congress can be obtained from 
the Collector of Internal Revenue of your district, or through the 
Congressman from your district. 

205. United States Income Taxes. — One schedule of 
rates is provided for the taxation of the incomes of 
individuals, and another for the incomes of corporations. 
The kind of income that is taxable is described in the 
revenue act. 

An Income Tax Return must be filed by every unmarried 
person whose net income for the year is more than $ 1000, 
and by every “ head of a family ” whose net income is more 
than $ 2500. For the former, $ 1000 is exempt from taxa¬ 
tion, and for the latter $ 2500 is exempt, and also for the 
latter $ 400 additional for each child under 18 years of age 
and for each dependent who is actually supported by the 
head of the family. 

The law provides for a Normal Tax and a Surtax on the 
net annual incomes of an individual. 

Normal Tax Rates. — On the first $ 4000 of net taxable 
income the rate is 4 % ; on the balance of the net income 
the rate is 8 %. 

Thus, a single man having a net income of $8500 is allowed an 
exemption of $ 1000, so that his taxable income is $7500. On the 
first $4000 of this, the tax is 4% or $160 ; on the remaining $3500 
the tax is Byb or $280. Hence his total normal tax is $440. 


TAXES 


299 


Surtax Rates. — If the net taxable income of an individual 
exceeds $ 6000, an additional tax is levied, called a Surtax. 
The rates differ according to the amount of the income. 


On All Income 

The Rate 
I s 

On All Income 

The Rate 
I s 

Between 

And 

Between 

And 

$ 6,000 

$10,000 

1% 

$100,000 

$ 150,000 

48% 

10,000 

12,000 

2% 

150,000 

200,000 

49% 

12,000 

14,000 

3% 

200,000 

300,000 

50% 

14,000 

16,000 

4% 

300,000 

400,000 

51% 

16,000 

18,000 

5% 

400,000 

500,000 

52% 

18,000 

20,000 

6% 

500,000 

750,000 

53% 

20,000 

22,000 

8% 

750,000 

1,000,000 

54% 

22,000 

24,000 

9% 

Over $1,000,000 

55% 

24,000 

26,000 

10% 




The rates continue to increase 




up to 47% on $100,000 





The following examples show how to compute the surtax 
and the total tax on larger incomes. 

Example 1. Find the total income tax of a man and wife 
who have one child, if their annual income is $ 18,000. 

Solution. 1. Their exemption is $2500 + $400, or $2900. 

2. .*. their taxable income is $18,000 — $2900, or $15,100. 

3. The normal tax: a. 4 °Jo of $4000 = $160 

h. 8 °Jo of $15,100 - $4000, or 8% of $11,100 = $888 

c. .*. the total normal tax = $160 4- $888, or $1048. 

4. The surtax: a. $6000 of $18,000 is exempt. 

b. 1% of $4000 between $6000 and $10,000 = $ 40 

c. 2% of $2000 between $10,000 and $12,000 = 40 

d. 3% of $2000 between $12,000 and $14,000 = 60 

e. 4 °fo of $2000 between $14,000 and $16,000 = 80 

/. 5 °}o of $2000 between $16,000 and $18,000 = 100 

g. .*. the total surtax — $620 

5. .*. the total tax = $1048 + $320, or $1368. 

















300 


NEW HIGH SCHOOL ARITHMETIC 


Example 2. What is the total tax of a man and wife, 
having two children, if their income is $ 25,000 ? 

Solution. 1. a. Their exemption = $2500. 

b. The exemption for two children is 2 x $400, or $800. 

c. Their total exemption is $2500 + $800, or $3300. 

d. .*. their taxable income is $25,000 — $3300, or $21,700. 

2. The normal tax: a. 4 °Jo on the first $4000 = $ 160 

b. 8 °Jo on $21,700 - $4000, or 8% on $17,700 = 1416 

c. .*. the total normal tax = $1576 

3. The surtax: a. $6000 of the $25,000 is exempt 

b. 1 °/o of $4000 between $6000 and $10,000 = 

c. 2% of $2000 between $10,000 and $12,000 = 

d. Z°Jo of $2000 between $12,000 and $14,000 = 

e. 4% of $2000 between $14,000 and $16,000 - 

f 5% of $2000 between $16,000 and $18,000 = 

g. 6% of $2000 between $18,000 and $20,000 = 

h. 8% of $2000 between $20,000 and $22,000 = 

i. §°fo of $2000 between $22,000 and $24,000 = 

j. 10 °Jo of $1000«between $24,000 and $25,000 = 

k. .*. the total surtax = 

4. .*. the total income tax = $1576 + $880, or $2456. 

EXERCISE 132 

Determine the total income tax under the law of 1932 for : 

l. A single man having an income of $ 3000 per year. 

2. A man and wife having an income of $ 5000 per year. 

3. A man and wife having one child, if their annual in¬ 
come is $ 3600. 

4. A man and wife having two children, if their annual 
income is $ 7500. 

5. A man and wife who have three dependents, if their 
annual income is $ 10,000. 

6. A man and wife having two children, if their income 
during 1932 was $ 20,000. 


$ 40 
40 
60 
80 
100 
120 
160 
180 
100 
$880 




TAXES 301 

7_12 - In each of the Examples 1 to 6, what per cent of 
the income is the income tax ? 

13. a. Suppose a single man has an income of $ 150,000. 
Compute his total income tax, making use of the fact that 
the surtax on $ 100,000 is $ 22,460. 

b. What per cent of his income is his income tax ? 

14. a . Suppose a man and wife who have four dependents 
have an income of $ 175,000. Compute their total income 
tax, making use of the fact that the surtax on $ 100,000 is 
$ 22,460. 

b. What per cent of their income is their income tax ? 

206. United States Inheritance Taxes at the present time 
(1933) are levied under the Revenue Acts of Congress of 
1926 and 1932. The total tax is the sum of the two taxes. 

$ 50,000 of an estate is exempt. The balance of the es- 
state is called the net estate. 


National Inheritance Tax Rates 


On That Amount op the Net Estate 

1926 Rates 

1932 Rates 

Total 

Up to $ 10,000 


1% 

1% 

2% 

Between $ 10,000 and $ 

20,000 

1% 

2% 

3% 

Between $ 20,000 and $ 

30,000 

1% 

3% 

4% 

Between $ 30,000 and $ 

40,000 

1% 

4% 

5% 

Between $ 40,000 and $ 

50,000 

1% 

5% 

6% 

Between $ 50,000 and $ 

100,000 

2% 

7% 

9% 

Between $100,000 and $ 

200,000 

3% 

9% 

12% 

Between $200,000 and $ 

400,000 

4% 

11% 

15% 

Between $400,000 and $ 

600,000 

5% 

13% 

18% 

Between $600,000 and $ 

800,000 

6% 

15% 

21% 

Between $800,000 and $1,000,000 

7% 

17% 

24% 

The rates continue to 

increase 




until they reach 45% on estates 
which have a net value of $10,000,- 




000 or more .... 


20% 

45% 

65% 













302 


NEW HIGH SCHOOL ARITHMETIC 


Example. What is the national inheritance tax on an 
estate having a net taxable value of $ 235,000 ? 


Solution. 1. 


Rate 

Tax 

The first 

$10,000 . . 

. . 2 % . 

. . . $200 

The second 

10,000 . . 

. . '6% . 

... 300 

The third 

10,000 . . 

. . 4 % . 

... 400 

The fourth 

10,000 . . 

. . 5% . 

... 500 

The fifth 

10,000 . . 

. . 6% . 

... 600 

The next 

50,000 . . 

. . 9 % . 

. . . 4,500 

The next 

100,000 . . 

. . 12% . 

. . . 12,000 

The remaining 

35,000 . . 

. . 15% . 

. . . 5,250 


Total Tax 


$23,750 


There are many provisions which cannot be explained in 
a book for high school pupils. 


EXERCISE 133 

Determine the national inheritance tax which must be 
paid out of an estate having a net value of: 


1 . 

$ 85,000 

5. 

$437,900 

2. 

$ 133,500 

6. 

$ 643,000 

3. 

$ 265,800 

7. 

$ 875,000 

4. 

$ 372,500 

8. 

$ 962,000 


9-16. In Examples 1-8, respectively, determine what per 
cent of the net value of the estate the national inheritance 
tax is. 









XII. BANKING 


207. Uses of Banks. Banks afford places of safe deposit 
for money. Money deposited in a checking account ordi¬ 
narily does not draw interest. Such deposits give the 
depositors a corresponding amount of Credit. 

Banks lend money or credit. 

Banks furnish convenient means of transferring money 
or credit from one person or firm to another. 

Promissory Notes 

208. A Promissory Note is a written promise of one per¬ 
son (or firm) to pay a specified sum of money to another 
specified person (or firm). 

The Maker of the note is the person who signs it, — the 
one promising to pay. 

The Payee of a note is the person to whom payment is 
promised. One of the following forms is used. 

a. Pay to John Davis. 

He alone may demand payment. 

h. Pay to John Davis or order. 

He or any one whom he designates may demand payment. 

c. Pay to bearer. 

Any one legally possessing the note may demand payment. 

The Face of a note is the sum to be paid. 

The place of payment is usually specified. If it is not, 
then the note is payable at the office of or else at the home 
of the maker. 

209. The Date of Maturity is the date when payment is 
due. 

a. A Demand Note is payable “on demand, 5 ' — payable 
303 


304 


NEW HIGH SCHOOL ARITHMETIC 


whenever payment is requested during business hours of a 
working day. 

b. A Time Note is payable at a future time that can he 
determined, — such as “ thirty days after date.” 

Note. — If payable a certain number of months after date, 
then calendar months are meant. 

If payable one month after the last day of any month, it is 
payable on the last day of the next month. 

If payable a certain number of days after date, the actual num¬ 
ber of days, including Sundays and holidays, is meant. 

If a note becomes payable on Sunday, or on a holiday, or on a 
Saturday when Saturday is a half-holiday, the note is payable on 
the next business day. 

Three additional days, called Days of Grace, in which 
payment might be made were formerly quite commonly 
allowed. In a few states they are still allowed. 


210. Interest-bearing Notes. — If interest is to be paid on 
the note, then the words “ with interest ” or “ with interest 
at a number of % ” are added after the place of payment. 

Value Received is commonly written on the face of the 
note, but is unnecessary. 


f. 6 ' 50 .. Newark, OUoJ‘AyJAi-19 3 .?. 

..fA-AfAt-Aftf..-.after date A.promise to pay to the 
order of 'Ufilltccms fkericl&'baxyyi, 

AA huncUU_ _ Dollars 

cA&umaJo cfatio-ncit fdccnk, 


with interest at^_ <Jo 
No. AAA. Due A A 


W, /<?3S ■■ 


/fewby 












BANKING 


305 


211. If a note is made payable to order or to bearer, it 
can be sold by the payee to another person. Such a note is 
Negotiable. Otherwise a note is Non-negotiable. The payee 
of a negotiable note may surrender his rights by Indorsing it 
on the back of the note, at its left end, in one of three forms. 


a 

b 

c 

William 

Henderson 


Pay to 

Martin Pratt 
or order 
William 
Henderson 


Pay to 
Martin Pratt 
or order 

without recourse. 
William 
Henderson 


The form a is called a “ blank indorsement ” ; form b a 
“ full indorsement ”; form c a “ qualified indorsement.” 
When using form a or 6 the indorser assumes responsibility 
for the payment of the note if the maker fails to pay. 
When using form c, the indorser does not guarantee the 
payment of the note. 

212. Joint and Several Notes. A note reading “We 
promise to pay ” and signed by two or more persons is a 
Joint Note. A note reading “ I promise to pay ” and signed 
by two or more persons is a Several Note. 

In either case, the payee may collect the whole payment 
if necessary from any one of the makers. When the note 
is a joint note, then each of the makers can hold the others 
for their share of the face of the note. 

213. To obtain payment of a note, the holder must pre¬ 
sent it to the maker on the date of maturity. If the maker 
refuses to make the payment, the holder may turn the note 
over to a notary public, who also presents it to the maker 
for payment. If payment is again refused, the notary 
makes formal Protest, and sends notice of the facts to all 
who have indorsed the note. Then, each indorser in turn 
becomes responsible for the payment of the note. 








306 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 134 

Write the following notes, dated to-day at your home town, 
payable at a fictitious bank: 


Ex. No. 

Face of 
Note 

Due 

Payee 

Maker 

Interest 

1 . 

$ 250 

In 3 mo. 

Henry Carroll 
or order 

Charles Darrow 

6 °Jo 

2. 

575 

In 90 da. 

The bank 

Stanley Fielding 

None 

3. 

625 

On demand 

The bank 

Frederick Burton 

5% 

4. 

1250 

In 60 da. 

James Davis 
or order 

Samuel Aldrich 

6 °/o 

5. 

1575 

In 6 mo. 

The bank 

Fred Allison 

6 °!o 


6. Write a blank indorsement by the payee for the note 
in Example 1. 

7. Write a full indorsement by the payee of the note in 
Ex. 4, transferring his rights to James Hanson, and the 
final indorsement by Hanson when he presented the note 
for payment at maturity. Also find the amount of the 
note at maturity. 

8-10. Find the amount at maturity of the notes in 
Examples 1, 2, and 5. 

214. In order to Guarantee the Payment of a Note the 

maker may: 

a . Get some other person with financial credit to in¬ 
dorse it; 

b. Offer security such as a mortgage on real property or 
certificates of ownership of other property. Such notes are 
called Collateral Notes. In form, they are similar to ordi¬ 
nary promissory notes except for additional clauses tempo¬ 
rarily transferring title to the property offered as collateral 
security. 













BANKING 


307 


Partial Payments 

215. Sometimes the holder of a note permits the maker 
of it to pay part of the face of the note before the date of 
maturity. Such payments are called Partial Payments. 

In many cases, the holder of a note will not accept any 
partial payments except on interest payment dates, and 
then only if the partial payment is a specified sum such as 
a multiple of $ 50. Thus, if a note is dated on June 5, 
with interest payable semi-annually, partial payments might 
be accepted only on Dec. 5, and June 5, and in multiples 
of $ 100. 

Whenever partial payments are accepted, they are recorded 
on the back of the note as below: 


t 8 . S .°. Newark, 19 6 

__ after date. f?..promise to 

, o itcL'yit&u o-'i <yicle/u 

pay to _ l - 

<oif6£ Dollars 

at__..with interest at. £7° 
Value received. 


jo-kfi 'Uiuv'i&n. 


Back of Note 


Received on this 
note: 

3/18/32 $200 

5/5/32 $150 

5/22/32 $300 

















308 


NEW HIGH SCHOOL ARITHMETIC 


216. One of two rules is employed to determine the 
amount still due on the note at maturity, when partial pay¬ 
ments have been made. 

The Merchant’s Rule is usually employed when final set¬ 
tlement is made within a year from the date of the note. 

Rule. — i. To the face of the note, add the interest upon it 
from the date of the note to the date of the settlement. 

2 . To each partial payment, add the interest upon it from the 
date when it was made to the date of settlement. 

3 . From the result of Step 1, subtract the sum of the results 
in Step 2. The remainder is the balance due. 

This rule will be applied to determine the amount due on 
Dec. 8,1932, on the note with indorsements shown in § 215. 

Solution. 1. The following form may be used to advantage. 


Note and 
Payments 

Date 

Made 

Date of 
Maturity 

Time 

Elapsed 

Face 

Interest 

on 

Amount 

of 

Note 

1 /2 / 32 

12/8/32 

11 mo. 6 da. 

$850 

$47.60 

$ 897.60 

Payments 

First 

3/18/32' 

12/8/32 

8 mo. 20 da. 

200 

8.67 

208.67 

Second 

5/ 5/32 

12/8/32 

7 mo. 3 da. 

150 

5.33 

155.33 

Third 

8/22/32 

12/8/32 

3 mo. 16 da. 

300 

5.30 

305.30 

Amount 

of all 
payments 






669.30 

Balance 
due on 

note 






228.30 


In columns 1, 2, and 4 are recorded the date of the note and of 
the payments, the date of final settlement, and the face of the note 
and of the payments. In column 3, is entered the difference in 
time between the corresponding dates in columns 1 and 2 ; thus, 
the first payment draws interest for 8 mo. and 20 da. The inter- 




























BANKING 


309 


est is then computed and entered in column 5, and the amount in 
column 6. Then the sum of the amounts of the three payments 
is found and subtracted from the amount of the note. This is the 
amount still due at the time of the final settlement. 


EXERCISE 135 

Using either the form suggested on p. 308 or some other 
neat form, determine the amount due at maturity on the 
following notes: 

1. A note for $900, with interest at 6%, dated March 
20 of this year and due October 20, has the following pay¬ 
ments on it: April 12, $200; July 18, $350. 

2. A demand note for $ 1050, with interest at 6 %, dated 
Jan. 8 of last year, was presented for settlement on Dec. 20. 
Partial payments as follows were recorded on it: March 15, 
$175; July 9, $325; Oct. 21, $250. 

3. A note for $ 2500, with interest at 6 %, dated Aug. 23 
of last year, was presented for settlement on May 12 of this 
year, after preliminary payments as follows : Nov. 17, $ 800; 
Jan. 25, $ 460 ; Feb. 16, $ 1200. 

4. On a note for $ 980 given May 18 of last year and 
drawing 5 % interest, three payments were made: June 10, 
$ 360; Sept. 3, $ 300; Oct. 17, $ 180. What was due 
April 27 of this year? 

5. On a note for $ 2600, dated Sept. 13 of last year and 
bearing interest at 6%, the following indorsements were 
made: Nov. 28, $300; Jan. 20, $450; March 12, $625; 
May 18, $ 520. How much was due July 28 of this year? 

6. A note for $ 1698, dated Jan. 14 of this year, is in¬ 
dorsed as follows: Feb. 27, $215; Aug. 15, $820; Aug. 28, 
$ 120; Oct. 11, $ 275. At 6 % interest, what is due 
Nov. 15 ? 


310 


NEW HIGH SCHOOL ARITHMETIC 


7. On a note for $3750, dated Nov. 17 of last year and 
bearing interest at 4 %, the following payments were made : 
Dec. 20, $1000; May 25, $550; June 21, $ 375; Sept. 22, 
$700. How much was due Oct. 27 of this year? 

8. A demand note for $ 1500, with interest at 6 °/ 0 , dated 
Feb. 24 of last year, was presented for settlement Feb. 8 of 
this year. Partial payments as follows were recorded on 
it: May 15, $150; Aug. 5, $175; Nov. 8, $225; Jan. 2, 
$ 325. How much was due at the time of settlement ? 

9. On a note for $ 3875, given July 2 of last year, and 
drawing 5 % interest, three payments were made: Sept. 5, 
$875; Dec. 1, $500; March 15, $375. What was due on 
June 15 of this year ? 

10 . A note for $ 2775, with interest at 6 %, dated Oct. 12 
of last year, was presented for settlement on Aug. 15 of this 
year, after the following payments had been made : Nov. 26, 
$175; Dec. 24, $225; Feb. 25, $125; April 25, $475; 
June 15, $ 775. How much was due ? 

217. The United States Rule. If the final settlement is 
made more than one year after the date of the note, the 
amount due is found by the following rule, adopted by the 
Supreme Court of the United States. 

Rule. — i. Find the amount of the face of the note to the 
time when the payment or the sum of the payments made 
equals or exceeds the accrued interest. 

2 . From this amount, subtract the payment or the sum of 
the payments. 

3 . Consider the remainder the new face of the note and pro¬ 
ceed as before up to the date of the settlement. 

Notice that the interest to date must be paid before any 
partial payment or payments can be used to decrease the 


BANKING 


311 


face of the note. The justification for this rule is that the 
lender has a right to expect his interest on the debt before 
consenting to apply a payment to a reduction of the debt. 

Example. What amount was due on Jan. 5, 1933, on a 
note for $ 850 dated Oct. 10, 1930, bearing interest at 6 %, 
on which the following indorsements had been made: Dec. 
23, 1930, $ 325; Aug. 11, 1931, $ 15 ; July 16, 1932, $ 150. 


Solution. 1. Face of note.1850.00 

2. Interest on $850 from Oct. 10, 1930, to Dec. 23, 

1930, 2 mo. 13 da. 10.34 

3. Amount to Dec. 23, 1930 .$860.34 

4. First partial payment. 325.00 

5. Subtracting, the new principal is.$535.34 

6. Interest on $535.34 from Dec. 23, 1930, to Aug. 

11, 1931, 7 mo. 19 da. 20.44 

(Since the second payment is less than this interest, 
no reduction of the principal or interest is made.) 

7. Interest on $535.34 from Aug. 11, 1931, to July 

16, 1932, 11 mo. 5 da. 29.89 

8. Amount of $535.34 to July 16, 1932 .$585.67 

9. The sum of the 2nd and 3rd partial payments . 165.00 

10. Subtracting, the new principal is.$420.67 

11. Interest on $420.67 from July 16, 1932, to Jan. 5, 

1933, 5 mo. 20 da.. • 1L92 

12. Amount due on Jan. 5, 1933 .$432.59 


EXERCISE 136 

Find the amount due on the following notes on the date 
of settlement, using the United States Rule. 

1. What amount was due July 17, 1934, on a note for 
$ 1395, dated Jan. 27, 1932, and bearing interest at 6 %, on 
which the following indorsements had been made: March 
15, 1933, $325; Oct. 25, 1933, $200. 

2. On Nov. 19, 1931, a demand note was issued for 
$680, with interest at 5 On Aug. 29, 1932, $20 was 
paid ; on Nov. 16,1932, $ 400. What was due Sept. 16,1933 ? 

















312 


NEW HIGH SCHOOL ARITHMETIC 


3. On a note for $2500, dated Feb. 21, 1929, and draw¬ 
ing interest at 6%, there was paid, Feb. 17, 1930, $850; 
July 12, 1931, $50; and Feb. 4, 1932, $750. What was 
due May 14, 1933? 

4. A note for $ 1750, dated Aug. 17, 1932, and drawing 
6 %, is indorsed as follows : April 19, 1933, $75; July 22, 
1933, $ 880; Nov. 12, 1933, $ 550. What was due Dec. 13, 
1933? 

5. A note for $540, dated April 16, 1930, and bearing 
interest at 4 %, had indorsements as follows: April 30, 
1930, $100; Nov. 26, 1930, $220; Aug. 13, 1932, $25. 
How much was due Nov. 29, 1933 ? 

6. On a note for $ 800, given May 25,1930, and bearing 
interest at 6 %, four payments were made: Feb. 13, 1931, 
$30; July 18,1931, $200; June 24, 1932, $24; Oct. 3, 

1932, $ 150. What was due Feb. 27, 1933 ? 

7. A note for $ 480, dated March 16,1932, is indorsed as 
follows: Aug. 29, 1932, $ 15; Sept. 12, 1932, $ 150; Sept. 
12, 1933, $ 120. At 7 % interest, how much was due April 
14, 1934? 

8. On a note for $ 1925, dated Dec. 12,1928, and bearing 
6 °/ 0 interest, the following payments were made: June 17, 
1929, $110; Sept. 23, 1930, $300; Nov. 11, 1931, $75; 
Oct. 17, 1932, $460; and March 27, 1933, $275. How 
much was due Aug. 21, 1934? 

9. A note for $ 575, dated July 14,1932, was indorsed as 
follows : Aug. 17, 1932, $ 50; Oct. 29,1932, $ 100; Jan. 15, 

1933, $ 130. How much was due on it on Sept. 15, 1933, if 
the rate of interest was 6 % ? 

10. The following payments were made on a note for 
$ 2340, bearing 6 % interest, dated Oct. 19, 1931: Jan. 15, 
1932, $ 250; July 10, 1932, $ 375; Dec. 20, 1932, $ 525; 
May 18,1933, $ 600. What was the sum due July 15,1933 ? 


BANKING 


313 


Discounting Notes 

218. Banks, when lending money, may deduct the inter¬ 
est in advance. 

Thus, a man borrowing $ 1000 from a bank for 60 days, with 
interest at 6 %, may be asked to pay in advance $10, the interest 
on $ 1000 at 6 °Jo for 60 days. 

The bank, in this case, is said to Discount the Note. 

A person or firm, having accepted from an individual his 
note, and wanting the money at once, may ask a bank to 
purchase the note. Again, the bank is said to Discount the 
Note. 

Example. The Acme Manufacturing Co. holds a non- 
interest-bearing note for $ 1000, payable in 60 days. They 
ask a bank to discount it. What are the proceeds ? 

Solution. The bank may offer $ 990 for the note. In doing 
so, the bank deducts $10 from the face of the note, this being 
the interest on $ 1000 at 6 °/o for 60 days. The proceeds are $ 990. 

219. Bank Discount is a charge made by a bank for the 
payment of a negotiable note before the note is due. 

The day on which the note is discounted may not be the 
day on which the note was written. The time from the day 
of discount to the date of maturity is the Term of Discount. 

The Discount is a certain per cent of the amount of the 
note at maturity. The rate per cent used is called the 
Rate of Discount. This may not he the same as the rate of 
interest payable on the face of the note. 

The Proceeds of a discounted note is the amount at matu¬ 
rity less the discount. 

Example 1. Kind the proceeds of a 90-day note for 
$500 discounted on the date of making at 6%. 

Solution. 1. Since the note is non-interest-bearing, its amount 
at maturity is $ 500. 


314 


NEW HIGH SCHOOL ARITHMETIC 


2. Being discounted on the date of making, the term of dis¬ 
count is 3 mo. 

3. The discount = the interest on $500 at 6 °Jo for 3 mo. == $7.50. 

4. .*. the proceeds = $500 — $7.50 = $ 492.50. 

Example 2. Henry Sanders asked his bank to discount 
for him on Oct. 25 of this year a note for $ 1200, payable 
in 90 days, without interest, dated Oct. 15 of this year. 
What were the proceeds ? 

Solution. 1. The note did not bear interest; hence the amount 
at maturity was $1200. 

2. Being a note for 90 days, it matures 90 days after Oct. 15. 
Hence on Oct. 25 it still has 80 days to run, — the term of dis¬ 
count. 

3. The bank offers to discount it at 6 °Jo. The discount, there¬ 
fore, is the interest at 6% for 80 days on $1200, or $16. 

4. The proceeds are $1200 — $16, or $1184. 

Example 3. Find the proceeds of a note for $ 750 dated 
May 12 of last year, payable 90 days after date, bearing in¬ 
terest at 5 %, if discounted June 23 of last year, at 6 %. 

Solution. 1. The interest on $750 at 5 °Jo for 90 days is $9.38. 
Hence the amount of the note at maturity =$ 750 + $9.38 = $759.38. 

2. The note matured 90 days after May 12, or on Aug. 10. 

3. The term of discount was from June 23 to Aug. 10, or 48 days. 

4. The discount was the interest on $759.38 for 48 days at 6</o, 
or $6.08. 

5. The proceeds were $759.38 — $6.08, or $753.30. 

EXERCISE 137 

Find the bank discount and the proceeds of the following 
notes, made last year. Assume that the rate of discount was 
6 % in each case. 

Each example provides material for three parts. 

In part a, the date of discount is the same as the date of 
the note; in parts b and c, the dates of discount are later 
than the date of the note. 


BANKING 


315 


If the teacher desires, part a for all may be done before parts b 
and c of the first example. Or, all three parts for examples 1, 2, 
and 3 may be done before any of the parts for the remaining ones. 
If additional examples are desired, the sum in any example may 
be changed to any other sum. 


Ex. 

No. 

Face of 
Note 

Date of 
Note 

Time of 
Note 

Rate of 
Interest 

Dat 

a 

E OF DlSCC 

b 

)UNT 

C 

1 . 

$600 

2/ 7 

60 da. 

none 

2/ 7 

2/20 

3/ 5 

2. 

250 

5/11 

90 da. 

none 

5/11 

6/ 1 

7/15 

3. 

875 

3/16 

6 mo. 

none 

3/16 

5/20 

7/ 2 

4. 

1000 

4/20 

30 da. 

6% 

4/20 

4/25 

5/ 1 

5. 

1600 

7/12 

20 da. 

6% 

7/12 

7/15 

7/25 

6. 

2400 

9/18 

18 da. 

5% 

9/18 

9/20 

9/30 

7. 

3000 

11/22 

15 da. 

5% 

11/22 

11/25 

11/30 

8. 

375 

7/29 

90 da. 

4% 

7/29 

8/30 

9/15 

9. 

560 

8/24 

1 yr. 

6% 

8/24 

12/20 

3/ 5 

10. 

1350 

12/15 

60 da. 

5% 

12/15 

1/10 

2/ 1 


11 . What are the proceeds of a 60-day note for $950, 
with interest at 5 %, discounted at 6 % ten days after the 
date of the note ? 

12. Find the bank discount on a note for $ 625, with in¬ 
terest at 6 %, payable in 4 months, if discounted 2 mo. 9 da. 
after the date of the note. 

13. What was the bank discount on a note for $ 850, dated 
July 13 of last year and payable, with interest at 6 %, 90 
days after date, if discounted at 6 % on Aug. 25 ? 

14 Wbat will be the proceeds of discounting at 7 % on 
the date of making a 30-day note for $3000, without 
interest ? 

15 Find the bank discount on a note for $ 910, dated 
Aug. 26 of last year, payable 90 days after date, with interest 
at 5 %, if discounted Nov. 3, at 6 
















316 


NEW HIGH SCHOOL ARITHMETIC 


16. Write a 90-day note for $738, of which yon are the 
payee and Henry Anthony is the maker, dated Aug. 27 of 
last year, with interest at 6%, payable at the Farmer’s 
Bank of Janesville, Ohio. Assume that you discounted 
the note on Sept. 15, at 6%, at some bank in your town. 
Properly indorse it over to that bank, and determine the 
proceeds. 

17. On Jan. 16 of last year, S. L. Payne made out in 
favor of the Anderson Automobile Co., of Anderson, Ky., 
the following notes to apply on the purchase price of an 
automobile: 

One for $ 250, due in 30 days, interest at 6 °/ 0 • 

One for $ 300, due in 60 days, interest at 6 %. 

One for $ 450, due in 90 days, interest at 6 %. 

a. Write the notes. 

b. On Jan. 20 the automobile company discounted all 
three notes at 6 %, at the Anderson State Bank, using the 
full form of indorsement in transferring them to the bank. 
Write the indorsements and determine the total proceeds of 
the notes. 

Find the proceeds of the following notes made last year: 


Ex. 

No. 

Face 

Date of 
Note 

Time of 
Note 

Rate of 
Interest 

Date of 
Discount 

Rate of 
Discount 

18. 

$475 

5/11 

30 da. 

6% 

5/20 

7% 

19. 

1250 

7/15 

90 da. 

5|% 

8/10 

6% 

20. 

360 

6/28 

45 da. 

8% 

7/15 

6% 

21. 

980 

10/20 

60 da. 

6% 

11/ 5 

7% 

22. 

125 

11/16 

15 da. 

8% 

11/20 

8% 

23. 

1450 

9/30 

2 mo. 

5% 

10/20 

6% 

24. 

3500 

8/24 

4 mo. 

54% 

10/15 

6% 

25. 

2675 

7/19 

6 mo. 

5% 

9/ 1 

6% 













BANKING 


317 


Exchange 

220. The process of transferring credit between indi¬ 
viduals and firms is called Exchange. 

a. Domestic Exchange is the process of transferring credit 
between individuals in this country. 

b. Foreign Exchange is the process of transferring credit 
between individuals in this country and those of another 
country. 

Exchange of credit is accomplished by means of Checks, 
Money Orders, and Drafts. 


Checks 

221. A Check is a written order upon a bank made by 
one of its depositors, to pay a specified sum of money to a 
specified individual or his order. 

The bank is the Drawee; the person signing the check 
is the Drawer ; the one to whom it is payable is the Payee; 
the sum to be paid is the Face of the Check. 


Buffalo, N. 

THE COMMERCIAL NATIONAL BANK 

Pay to the order of..... _ IffJi 1 

and 8/[ 

Id^Cltdamv (X. Tyia'ik^. 


Henry Sears may cash the check himself or he may sur¬ 
render his right by using one of the three forms of indorse 
ment discussed in § 211. 








318 


NEW HIGH SCHOOL ARITHMETIC 


222. If Henry Sears as well as William Marks is a de¬ 
positor of the Commercial National Bank, and Henry Sears 
presents the check for deposit at this bank, the clerks there 
merely transfer credit of $ 25.31 from the account of Mr. 
Marks to the account of Mr. Sears. This is the simplest 
form of the problem of exchange. 

Mr. Sears, or some one to whom he indorses the check, 
may deposit or cash the check in a different bank. 

Banks are in this way daily cashing or accepting for de¬ 
posit checks drawn upon other banks. In a small commu¬ 
nity, each bank is likely to send a messenger daily to each 
of the other banks of the community whose checks have 
been accepted, to ask for money to the value of these checks. 
Thus, credits in one bank are transferred to another bank. 
This also is a simple problem of exchange. 

In a larger community, the settling up between banks 
is accomplished through a Clearing House. Clerks repre¬ 
senting each of the members of the clearing house associa¬ 
tion gather at a specified time, exchange with one another 
the checks of their respective banks, recording the amounts 
thereof, and report to the manager of the association the 
totals of these exchanges. By suitable procedure, the in¬ 
debtednesses between the member banks are all paid within 
a few hours. This is the most complex problem of exchange 
in a given community. 

223. Banks in small communities usually have funds on 
deposit in at least one bank of a near-by larger community. 
In particular, banks of the country are grouped in twelve 
districts in each of which is a Federal Reserve Bank, with 
which most of the banks of that district are associated. 

224. Payments made by Checks to persons in other towns. 

Example. A man in Texas may send a check to his son 

m college at Cambridge, Massachusetts. His son may deposit 
the check in a Cambridge bank. The bank will send the 
check, probably, to the Federal Reserve Bank at Boston, 


BANKING 


319 


receiving credit for it. The Federal Reserve Bank at Bos¬ 
ton will send it to the Federal Reserve Bank at Dallas, 
Texas, receiving credit for it. If the bank on which the 
check is drawn is a member of the Dallas Federal Reserve 
Bank, the latter, deducting the amount of the check from 
the account of the issuing bank, sends the canceled check 
to that bank. The issuing bank, in turn, deducts the amount 
of the check from the balance of the drawer. Thus, credit 
has been transferred from the father in Texas to the son in 
Massachusetts. Sometimes a small fee, % of the face 
of the check, is charged as exchange by the bank cashing 
the check. The other banks handle the check as a matter 
of business courtesy between themselves. 

This process represents a complicated form of Domestic 
Exchange. 


Other Ways of Transferring Credit 


225. Since a check is signed only by an individual, per¬ 
sons unacquainted with him, or the payee, often do not wish 
to accept the check. Hence, for transfers of funds to a 
considerable distance certain other means are often employed. 

226. A Certified Check is a personal check of a depositor 
in a bank, across the face of which an officer of the bank 
has written over his signature the word “ Good ” or 
“ Certified.” 


Bridgeport, Coring -- 



THE CONTINENTAL BANK 



t 65 Wo 


Pay to the order of- _ 
o futety-ftw-e and j~. 


Dollars 


TyiaAXJva CCLd'bi&h, 







320 


NEW HIGH SCHOOL ARITHMETIC 


Such certification dispels any doubt about the value of 
the check, for it is a criminal offense for a bank to certify 
a check unless the face of it is actually on deposit in the 
bank. 


227. A Cashier’s Check is a check drawn by the cashier 
of a bank, making a sum of money payable to an individual. 
There is not any charge, usually, for issuing such a check 
to a regular depositor of a bank. 


Moline, III., - - _ 3*1 3. _ _ _ 19?A 

THE TRADERS NATIONAL BANK 

Pay to the order of..3^^333^ JJ1 S 3. 

of&v-&Pb hwyuiUs&oC yvinety-fiv-e avid 

ScLw-avoL <9cuz& 

C ashier s Check --- 


To obtain such a check, Mr. Winters must pay the 
bank the equivalent of it. The check is readily accepted 
in exchange, for it has behind it the reputation of the bank. 

228. There are other methods of transferring money 
from one community to another, which should be men¬ 
tioned because of their connection with the preceding 
sections. 

Post Office Money Orders may be used to transfer credit 
either to a point in this county or abroad. They are espe¬ 
cially useful for sending small sums of money. 

The members of the class should each obtain or be given 
an Application for Domestic Money Order. They may be 
obtained at any post office. Upon the back of this Appli¬ 
cation will be found the rates for money orders. 







BANKING 


321 


Money may be Transferred by Telegraph. The only ad¬ 
vantage in so transferring credit is the speed with which 
it can be done. The rates and provisions are obtainable at 
any telegraph office. 

Travelers’ Checks are nsed very commonly by persons 
who are traveling. Express companies and banks charge 
\°/o of the face of the check as commission. 


EXERCISE 138 


What is the rate for a money order for: 


1. 

75/? 

4. 

$ 11.32 ? 

7. 

$ 2.78 ? 

10. 

$67.84? 

2. 

$ 1.65 ? 

5. 

$ 9.45 ? 

8. 

$ 6.32 ? 

li. 

$ 43.26 ? 

3. 

$ 4.98 ? 

6. 

$ 22.63 ? 

9. 

$ 35.69 ? 

12. 

$72.58? 

What will be the charge for a traveler’s check for : 

13. 

$200? 

16. 

$ 1500 ? 

19. 

$675? 

22. 

$225? 

14. 

$350? 

17. 

$ 2500 ? 

20. 

$3200? 

23. 

$385? 

15. 

$575? 

18. 

$ 1850 ? 

21. 

$1250? 

24. 

$750? 


229. A Certificate of Deposit also may be nsed in trans¬ 
ferring credit. 


THE COMMERCIAL BANK 




Detroit, Mich., __*??'_ 19 

__ has deposited 

f/'W-elv-e hw'vicL'b&cL Dollars 

payable to the order of --- 

on the return of this certificate properly indorsed. 

/ferny ffo-w-ne 

Cashier 








322 


NEW HIGH SCHOOL ARITHMETIC 


Such certificates of deposit serve as receipts for funds 
left on deposit with a bank. Banks pay interest on such 
funds. 

In one community, 2\ % interest is allowed on funds 
remaining on deposit three months, 3 % on money on de¬ 
posit six months, and 4 % on money on deposit one year. 

By properly indorsing the certificate, the owner of it 
may transfer his credit to some one else. 

The certificate may be made payable to some one else at 
the start. 


EXERCISE 139 

1. a. Write a certificate of deposit, dated to-day from a 
fictitious bank in your town, stating that you have deposited 
$ 350 payable to yourself. 

b. Assume that you leave the money on deposit three 
months. What interest will you draw ? 

c. Assume that you wish to transfer the money at the 
close of the three months to Henry Randolph. Write the 
proper indorsement. 

2. What interest at the rates mentioned above will be 
due on a certificate of deposit: 

a. Eor $ 1800 left on deposit six months ? 

b. For $ 785 left on deposit three months ? 

c. For $ 1235 left on deposit for one year ? 


Drafts and Acceptances 

230. A more popular means of sending money to a dis¬ 
tance is a special form of check called a Draft. 

A Draft is a written order from one individual, firm, or 
bank to another, directing the latter to pay a specified sum 
of money to a third party. 


BANKING 


323 


The Drawer is the person signing the draft; the Drawee 
is the person, firm, or bank to whom the draft is directed; 
the Payee is the one to whom payment is to be made. 

Example. Assume that Mr. Sampson of Philadelphia, 
Pa., owes Mr. Bristol of Peoria, Ill., $ 600. He may go 
to a bank in Philadelphia and purchase a draft, payable to 
Mr. Bristol. Since the Philadelphia bank does not have a 
correspondent in Peoria, it will probably give him a draft 
on a Chicago bank. 

To obtain this draft, Mr. Sampson must pay the Phila¬ 
delphia bank $ 600. Also, a charge of % of the face of 
the draft is often made. This charge is very commonly 
omitted when the purchaser of the draft (Mr. Sampson) is 
a regular depositor in the bank. 

Mr. Sampson will send the draft to Mr. Bristol. The 
latter, to obtain the money, will proceed exactly as if he 
had received a check (§ 221). The bank cashing the draft 
may make a charge for the cost of collecting the face of the 
draft. 


Note. — 1. Mr. Sampson may have the draft made payable to 
him. In this case, he will indorse it over to Mr. Bristol, using 
the full indorsement (§ 210) and then send it to Mr. Bristol. 

Note. — 2. The words “value received” and “charge to the 
account of ” are not necessary. 

When both the drawee and drawer are banks, the draft 
is called a Bank Draft. A bank draft is particularly desir¬ 
able as a means of sending money to a distance, because it 
has behind it the business reputation of a bank. 

Usually, a bank draft is a sight or demand draft. Pay¬ 
ment is specified by such phrases as “ pay on demand or 
“ pay at sight.” It is payable as soon as presented and will 
be cashed or accepted for collection by any bank. 


324 


NEW HIGH SCHOOL ARITHMETIC 


EXERCISE 140 

Write the following drafts, and determine the charge if 
the rate in each case is % ; date each to-day. 


Ex. 

No. 

Drawer 

Drawee 

Payee 

Amount 

1 . 

Farmer’s Bank, 

First N ational Bank, 

Charles 



Canton, Ohio 

Cleveland, Ohio 

Hancock 

$325 

2. 

A fictitious bank 

Exchange Bank, 

Sarles and 



in your town 

Detroit, Mich. 

Davis 

674.45 

3. 

A fictitious bank 

Continental Bank, 

Marshall Field 



in your town 

Chicago, Ill. 

and Co. 

835.60 


4. Charles Johnson, Ames, Pa., owed William Jordan of 
Pittsburgh, Pa., $ 175. He applied at his bank, the Ames 
Merchants Bank, for a sight draft payable to William Jor¬ 
dan. The Merchants Bank through their cashier, Arthur 
Giles, drew upon the Second National Bank of Pittsburgh 
on Peb. 14, 1933. Write the draft. 

5. Using the same data as in Ex. 4, make the draft 
payable to Charles Johnson and then write a full indorse¬ 
ment of it by Johnson over to William Jordan. 

231. Drafts are used also as a Means of Collecting Money 

in commercial transactions. Such drafts are called Com¬ 
mercial Drafts. They may be either sight drafts or time 
drafts. 

Commercial Sight Draft. Suppose that the Northern Mill¬ 
ing Company of Minneapolis, Minn., sells to Henry Marston of 
Maumee, Ohio, a carload of bran. They receive from the rail¬ 
road company a Bill of Lading. Mr. Marston must present 
this bill of lading to the railroad delivering the car before 
he can get possession of the car. If the Milling Company 
wants to insure payment for the bran, they make out the 
following sight draft upon Mr. Marston. 














BANKING 


325 


™ Minneapolis, Minn., ___l#f£ 

Cit Nqkt , 

- L -ytraz/ the order 

oj? <Tk& Jyiate'ia/ &9C@Jva, / yi(f& Banks TYlLvin&a.f&o-lLa, 


okcnov h'lMoUseoL &Lqktu-iLv-& _ _ _ *2_ ,, 

- 1. — f-E - C^EEC^^Eiand 100 Dollars 

Value received and charge the same to the account of 

The Northern Milling Co • 

UMlUm, /-foAjUtt, 

Tklaunv&e, (fokto-. Secretary. 


The Milling Company deposits this draft, attached to 
the hill of lading, at its hank in Minneapolis. The hank 
sends it to a hank in Maumee, Ohio. That hank notifies 
Mr. Marston that it has the hill of lading and draft. Mr. 
Marston, after paying the face of the draft, receives the bill 
of lading, and can then get his carload of bran. 

The Maumee hank may charge a small fee for its services. 
This charge may be a small fee (25 or a small per cent 
(_i_ %) of the face of the draft. It sends the net proceeds 
to the Minneapolis hank, and the latter credits the account 
of the Milling Company. The Minneapolis hank may, hut 
probably does not, make any charge for its services to the 
Milling Company, because the latter are regular customers 
of the bank. The minimum charge is usually 25 f 

232. A Commercial Time Draft is payable after the expira¬ 
tion of a specified length of time. Payment is specified by 
such phrases as “ 15 days after date,” or “ 15 days after 
sight.” 

In such cases, the draft is presented to the drawee to 
determine whether he intends to pay the money. If the 













326 


NEW HIGH SCHOOL ARITHMETIC 


drawee decides to pay the draft, he signifies his intention 
by writing across the face of the draft the word “ Accepted ” 
with the date and his name. 

After the draft is accepted, it is called an Acceptance. 

An acceptance is a form of promissory note. The drawee, 
having accepted it, thereby promises to make payment at 
a certain time. 

The date of maturity is reckoned as the specified number of 
days from the date of the draft , if it is payable a number of 
days after date , or as the specified number of days from the date 
of acceptance if it is payable a number of days after sight. 

Example. Assume that Henry Green of Windsor, Ill., 
has owed Arthur Brown of Boston, Mass., $ 250 for some 
time, and that Brown wishes to compel Green to pay off 
the debt. Brown may have his bank draw up the follow¬ 
ing draft. 


$250 


Boston, Mass. > ..F2F'.F.L..19?% 




order of. _ 

cTW-o 4 'wncUseoL 



ftawv&Q, CLcLayyy^ 
Cashier 


W vncl&cyi, JIL 


Mr. Brown deposits the 
draft in his bank, indors¬ 
ing it as at the right. 


Pay to 

the order of the 
First National Bank 
for collection only. 
Arthur Brown. 










BANKING 


327 


The First National Bank 
sends the draft to the 
Windsor State Bank of 
Windsor, Ill., for collec¬ 
tion, indorsing it as at the 
right. 


Pay to the order of 
The Windsor State Bank 
Windsor, III. 
for collection only. 

The First National Bank 
Boston, Mass. 


The Windsor State Bank informs Mr. Green that it has 
a draft drawn npon him. He, wishing to retain a good 
business reputation with the bank, decides to pay the 
draft; he writes across the face of it his “ acceptance,” as 
indicated on the draft, by writing across its face the word 
“ Accepted,” the date, and his name. 

Having accepted it on April 20,1932, he must pay it on 
or before June 19, 1932. 

When Green pays the draft, his bank marks it as paid 
and delivers it to him; also, after deducting, possibly, a 
small fee such as Jg- % of the face for collection, his bank 
remits the net proceeds to the First National Bank. The 
latter places the proceeds to the credit of Mr. Brown. 

EXERCISE 141 

1. a. Write a thirty day time draft of which you are the 
payee, and also the drawer, and of which Samuel Hughes, 
Pittston, Pa., is the drawee. Date it to-day. Assume 
that Mr. Hughes accepts the draft five days from to-day. 

b. Write across the face of the draft his acceptance. 

c. What is the date of maturity ? 

2. a. Write a sixty-day time draft, dated to-day, at your 
town, of which you are the payee and also the drawer, drawn 
upon the Paducah, Tenn., Milling Company, b. Assume 
that you deposit it for collection at a fictitious bank in 
your town. Properly indorse it. c. Assume that your 
bank sends it for collection to a fictitious bank in Paducah 





328 


NEW HIGH SCHOOL ARITHMETIC 


Properly indorse it for the bank. d. Assume that the bank 
presents it to the Milling Company and that the company 
accepts it ten days from to-day. Write the acceptance. 
e. What is the date of maturity ? /. Suppose that the 
Paducah bank charges ro °Io for collection. What will be 
the proceeds ? 

3. Ernest Waters, Williamstown, Ky., bought $850 
worth of merchandise from Dickson, Hughes, and Com¬ 
pany, of Toledo, Ohio. The latter, after waiting some time 
for their money, made out on July 25, 1932, a thirty-day 
time draft, payable to themselves, naming Mr. Waters as 
drawee. They deposited the draft for collection with the 
Republic Bank of Toledo, Ohio. The latter sent it to the 
Center Co. National Bank at Williamstown, Ky., for pres¬ 
entation. On July 30, Mr. Waters accepted it. 

a. Write the draft, the indorsements, and the acceptance. 

b. When is the draft due ? 

c. If the Williamstown Bank makes a charge of To ° 1 <> 
for collection, what amount will they remit to the Toledo 
Bank? 

233. In case the payments from one community to an¬ 
other are much greater than the collections for it from the 
other, then it becomes necessary to send actual money from 
the first community to the second to pay the indebtedness. 
Persons in the first community asking for drafts to make 
payments in the second community will then be asked to 
pay a fee for exchange, amounting possibly to 1 % of the 
face of the draft. In this case, Exchange on the second 
community is at a Premium or above Par. 

On the other hand, if the credits of one community are 
greater than its debits, then the person offering to purchase 
a draft payable in the second community may be offered 
a discount. Exchange is then at a Discount or Below Par. 


BANKING 


329 


The Rate of Exchange is the rate per cent of the face of a 
draft charged for drawing the draft when business con¬ 
ditions make it necessary to control the flow of money from 
or into a community. This charge is in addition to the fee 
for issuing the draft. 


EXERCISE 142 

1. Is the flow of money encouraged or discouraged when 
exchange is above par ? Below par ? 

Kind the cost of a sight draft for: 

2. $500 when exchange is at a premium of —no 
charges for issuing the draft. 

3. $285 when exchange is at a discount of f %, if the 
charge for issuing the draft is Jy %. 

4. $1975 when exchange is at a discount of — 
charges y 1 ^ % for issuing the draft. 

5. $ 2500 when exchange is at a premium of % , — no 
charge for issuing the draft. 

6 . $ 650 when exchange is at a discount of \ %,— tu% 
for issuing the draft. 

234. Discounting Acceptances. Since an acceptance is a 
form of promissory note, it will be discounted on the same 
conditions as a promissory note. It will be assumed that 
the drafts are payable a number of days after sight, unless 
something is said to the contrary. 

Example. The Hartford Manufacturing Co. had a ninety- 
day draft for $ 1250 drawn upon James Markley, dated 
May 12, and accepted by Markley on May 15. Needing 
the money, they asked their bank to discount the accept¬ 
ance on May 25. What were the proceeds if the rate of 
discount was 6 % ? 

Solution. 1. The date of acceptance was May 15. 

2. .\ the date of maturity was 90 days after May 15. 


330 


NEW HIGH SCHOOL ARITHMETIC 


3. The acceptance was presented 10 days after May 15; hence 
the term of discount (§218) was 90 — 10, or 80 days. 

4. the discount was the interest on $1250 at .6 °Jo for 80 
days. This is $16§, or $16.67. 

5. .\ the proceeds were $1250 — $16.67, or $1233.33. 


EXERCISE 143 

Eind the proceeds of the following acceptances, made last 
year, if discounted at 6% : 


Ex. 

No. 

Face 

Date 

Time 

Accepted 

Discounted 

a 

b 

1 . 

$300 

7/14 

30 da. 

7/15 

7/15 

7/20 

2. 

2850 

10/ 3 

90 da. 

10/10 

10/10 

11/ 1 

3. 

625 

9/24 

60 da. 

9/30 

9/30 

10/25 

4. 

1340 

11/ 8 

30 da. 

11/10 

11/10 

11/15 

5. 

932 

10/17 

20 da. 

10/20 

10/20 

11/ 1 


6. Assume that the Brown Hardware Co. of Middlesex, 
Delaware, received a statement dated Jan. 5, 1933, from the 
Lamson Manufacturing Co. of New York City, of an over¬ 
due account amounting to $ 626.50, with the request that it 
be paid at once or that the Brown Hardware Co. accept 
a 30-day time draft, payable to the Lamson Manufactur¬ 
ing Co. 

a. Write the draft, making the Lamson Co. both payee 
and drawer. 

b. Indicate upon it the Brown Hardware Co.’s acceptance 
under date of Jan. 7, 1933. 

c. Assume that the Lamson Manufacturing Co. discounted 
it on Jan. 10, 1933, at the First National Bank of New 
York. Properly indorse the acceptance, and determine the 
proceeds if the rate of discount was 6 %. 














BANKING 


331 


7. Dr. Theodore Daniels, Athens, So. Carolina, demanding 
payment of Mr. Sommers on an overdue bill amounting to 
S135, addressed to him on Aug. 19, 1933, a 60-day time 
draft for that amount payable to the doctor. 

a. Write the draft. 

b. Indicate Mr. Sommers 7 acceptance on Aug. 25, 1933. 

c. On Oct. 1, 1933, Dr. Daniels discounted the acceptance 
with Stephen Andrews at 6 %, making out a full indorse¬ 
ment of it to the latter. Write the indorsement, and deter- 
mine the proceeds. 

8. On Nov. 25,1933, the Commoners Bank of Watertown, 
New Jersey, discounted for customers the following com¬ 
mercial paper: 

1. A 3-mo. note for $ 1725 at 6 % dated Nov. 10. 

2. A 30-day time acceptance for $ 7235 dated Nov. 12, 
and accepted Nov. 15. 

3. A 60-day non-interest-bearing note for $ 800 dated 
Nov. 25. 

4. A 90-day note for $ 2375 at 6 % dated Oct. 18. 

5. A 30-day acceptance for $ 10,000 dated Nov. 18, and 
accepted Nov. 19. 

a. Find the amount the bank paid for each of these 
securities. 

b. On Nov. 30 the bank re-discounted all these notes at 
the National Bank of New York. Find the amount credited 
to their account by the latter bank. 

Foreign Exchange 

235. A draft used in Foreign Exchange is always called a 
Bill of Exchange. Formerly, when ocean transportation 
was slow, and less certain, bills of exchange were always 
issued in triplicate. These were sent by different routes 
and the first to arrive was paid, the others becoming void 


332 


NEW HIGH SCHOOL ARITHMETIC 


thereby. Now, however, they are issued at most only in 
duplicate, one copy being placed on file as a matter of 
record in case the other is lost. When issued in duplicate, 
they take the following form : 


V 50 Boston, 

t dt sight of this First of Exchange ( Second, of the 
same date and tenor, unpaid) pay to the order of 

Fwo- hwncL'hecl cPo-uricL^ 

Value received and charge to the account of 
c To- Thl&b&ul. JSuVWVLCf fSvo-b. 

ATwMer, Peabody and Co. 


Exchange may be transmitted also by cable. 


Lo-ncio-'K, 


236. Rates of Foreign Exchange depend upon the balance 
in trade between this country and the foreign country to 
which it is desired to send money. Exchange on foreign 
countries is quoted daily in the newspapers of large cities. 

In exchange between the United States and England, 
£ 100 is theoretically equivalent to $ 486.65, and $ 100 is 
theoretically equivalent to 518.1 fr. in exchange between 
the United States and France. The following table shows 
the rates of exchange on a certain day. 


London 


Cables 476.425 
Checks 475.275 


France 


Cables 570 
Checks 571.5 


This means that £ 100 was equal to $ 476.42^ if ex¬ 
changed by cable, and to $ 475.275 if exchanged by draft'; 
that $ 100 was equal to 570 francs if exchanged by cable, 
and to 571.5 if exchanged by draft. 









BANKING 


333 


Comparison with, the theoretical equivalents shows that 
exchange with England and Erance was below par on that 
day, indicating that the balance of trade was in favor of the 
United States, since bankers deemed it necessary to encour¬ 
age shipment of funds abroad by giving more than the theo¬ 
retical equivalent. 

Example. What was the cost of cable exchange on 
London for £ 160 8s. when exchange was quoted at 476.425 ? 

Solution. 1. £ 160 8s. = £ 160.4. 

2. Since each 1£ cost $ 4.76425, then £160.4 cost 160.4 x 
$4.76425, or $764.19. 

Example 2. Find the cost of a bill of exchange on France 
for 632 fr. when exchange was quoted at 570. 

Solution. 1. $100 was considered the equivalent of 570 fr., or 
$1 of 5.7 fr. 

-2. .\ the cost of 632 fr. was as many dollars as 5.7 is contained 
times in 632, or $ 110.87. 

EXERCISE 144 

What will be the cost of a bill of exchange: 

1. On London for £257 when exchange is quoted at 
485? 

2. On Paris for 1750 fr. when exchange is quoted at 
524? 

3. On London for £ 3525 when exchange is quoted at 
484.75 ? 

4 . On Paris for 4275 fr. when exchange is quoted at 
516.25 ? 

5. On London for £ 8500 when exchange is quoted at 
485.875 ? 

6. How large a bill of exchange on London can be 
bought for $ 2500 when exchange is quoted at 486 ? 


334 


NEW HIGH SCHOOL ARITHMETIC 


7. A family wishes to send $ 1000 to a member in 
Paris. If exchange is quoted at 518, how many francs can 
be obtained ? 

8. Find the cost of a bill of exchange for £ 750 10s. 
when exchange is quoted at 485.5. 

9. An importer wants to send his purchasing agent in 
London $ 5000. What will be the face of the bill of ex¬ 
change if exchange is quoted at 484.5 ? 

10. The United States Government loaned the French 
Government $ 100,000,000 at one time during the month of 
November in 1918. If exchange was quoted at 571.5, what 
was the equivalent in francs ? 


XIII. PARTNERSHIPS AND CORPORATIONS 


237. A Partnership is an association of two or more per- 
sons, called Partners, formed by a written contract, to 
engage in business in common for the purpose of making 
profits. 

Note. — Students will find an interesting article discussing the 
subject of Partnership in the Encyclopaedia Britannica. 

Partnerships are organized under special laws of the 
various states. 

The total money and property contributed by the partners 
is called the Capital. 

The Assets of a company comprise all its money and 
property at any particular time, and the Liabilities of a 
company are its debts. 

238. A General Partner may be a real partner, one who 
actually shares in the management and profits of the busi¬ 
ness ; a silent partner, one who shares in the profits but 
not in the management of the business; an ostensible 
partner, one who appears to be but is not actually a partner. 

The general partners of a business are individually re¬ 
sponsible for all liabilities in excess of the assets. 

In some states, special partnerships may be formed. In 
such, there must be at least one general partner. The 
special partners contribute to the capital, and are liable for 
debts of the company only to the extent of their contribu¬ 
tion to the capital. 

A partnership is sometimes for a fixed term of years and 
sometimes for an indefinite period. A partner cannot 
withdraw from a firm before the close of the term of 
agreement unless the partnership is dissolved. 

335 


336 


NEW HIGH SCHOOL ARITHMETIC 


The death of a partner, the bankruptcy, or the insanity 
of a partner automatically bring about the dissolution of 
the firm. A partner cannot dispose of his interests with¬ 
out the consent of the other partners. 

The manner of distributing the profits is specified in the 
contract between the partners. In the following examples 
some methods of distribution are described. 

EXERCISE 145 

1. Andrews, Baldwin, and Davis form a partnership, 
each contributing $ 3000, and each contributing services of 
equal worth. They agree to divide the profits equally. If 
the net profits are $ 3780, what should each receive ? 

2. A and B formed a partnership, A contributing $ 2000 
and B $ 4000. They gained $ 2460. They agreed to divide 
the profits in proportion to their respective contributions to 
the capital. 

Suggestions. — 1. What was the total capital? 2. What part 
of the capital did A contribute? 3. What part of the profits 
should he receive*? 

3. A, B, and C form a partnership to which each con¬ 
tributes equal services; A contributes $ 2000, B $ 3200, 
and C $ 4800. They agree to divide the profits in propor¬ 
tion to each partner’s contribution to the capital. What 
should each partner receive out of $ 4350 profits ? 

4. A, B, and C form a partnership. A contributes 
$ 8000, B $ 7500, and C $ 6500. They agree to divide the 
profits in proportion to their shares of the capital. How 
much should each receive out of the $ 6275 profits ? 

5. Suppose that, in the preceding example, the partners 
also agree to divide any losses as they divide the profits. 
If the assets of the company at the time of dissolution are 
$ 22,750, and the liabilities are $ 28,430, how much should 
each contribute to pay off the indebtedness ? 


PARTNERSHIPS AND CORPORATIONS 337 


6. Assume that partner A in Example 5 is bankrupt at 
the time of dissolution. How much must B and C each 
pay if they agree: a. To pay their own shares and each 
to pay one half of A ? s share ? b. To pay the debts in pro¬ 
portion to their contributions to the capital ? 

7. Assume that partners A and B in Example 5 are both 
bankrupt at the time of dissolution and that C has great 
wealth. How much must C pay to the creditors to close 
up the affairs of the firm ? 

8. A and B form a partnership. A contributes $15,000 
and B $ 5000 to the capital. They agree that A is to re¬ 
ceive from the business a salary of $ 150 per month and 
B $ 100 per month, and that the net profits at the end of 
the year shall be divided in proportion to their contribu¬ 
tions to the capital. The average monthly income is $ 450, 
from which the salaries are deducted. What should each 
receive at the end of the year from the accumulated net 
profits ? 

9. A and B form a partnership as follows: A contrib¬ 
utes $ 7300 and B, $4700. They agree that each shall 
receive at the end of the year 6 °Jo interest on his share of 
the capital, and that they will divide equally the surplus 
profits. What should each receive out of $2350 gross 
profits ? 

10. Suppose that A and B in Example 9 agree that the 
surplus profits, after paying the interest on invested capital, 
shall be divided in proportion to the amount of capital in¬ 
vested. What should each receive in this case ? 

11. The following plan is a very common one. 

A and B invest $ 6000 and $ 10,000 respectively in a 
business. They agree that A is to receive from the busi¬ 
ness annually $ 1500 and B, $ 1800; that each is to receive 
6 °J 0 interest on his invested capital; and that the surplus 
profits are to be divided equally. What does each receive 
out of gross profits of $ 5200 ? 


338 


NEW HIGH SCHOOL ARITHMETIC 


12. Assume that A, B, and C contribute $ 6300, $ 520Q 
and $ 7500, respectively, to the capital of a company; that 
they agree to allow each a salary of $ 125 per month; that 
each is to receive 6 % interest on his investment; and that 
surplus profits at the end of the year are to be divided 
equally. What should each receive during the year out of 
average monthly profits of $ 925 ? 

13. A, B, and C formed a partnership, agreeing that each 
should receive from the profits 6 °J 0 interest on his invested 
capital, and that surplus profits should be divided equally. 
Each invested $ 2500 on Jan. 2. A invested $ 500 on May 1, 
and $ 400 on Aug. 15. B invested $ 200 on March 10, 
$ 100 on July 20, and $ 150 on Sept. 1. 

The gross profits at the end of the year were $2350. 
How much should each partner receive ? 

14. A, B, and C contributed to the capital of their firm 
$ 1200, $ 3300, and $ 2700, respectively. They agree that 
each is to receive 6 °J 0 interest on his invested capital and 
that the net profits shall be divided equally. At the close 
of the year (Dec. 31), they found that they had liabilities 
as follows : bills payable, $ 736.45 ; a note for $ 500, dated 
Nov. 20, with interest at 6 %, which they must pay on or 
before Jan. 30. 

They have the following assets : cash on hand, $ 1638.75 ; 
bills receivable, $1482.31; a 90-day note for $375, with 
interest at 6 %, dated Dec. 1; a 30-day acceptance for $ 250, 
dated Dec. 15 ; a 60-day acceptance for $ 425, dated Nov. 25. 

a. Assume that they discount at their bank at 6 % on 
Jan. 2 the note payable to them and the acceptances, add¬ 
ing the proceeds to the cash on hand. What is the total 
amount of cash on hand then ? 

b. What is the amount of the note which they owe 
(principal + interest to date) ? Assume that they pay this 
amount and also all their outstanding bills. How much 
cash on hand remains ? 


PARTNERSHIPS AND CORPORATIONS 339 


c. Out of their remaining cash assets they pay to each 
partner the interest on his investment. What does each 
receive ? 

d. What is the balance of their book profits, including 
cash on hand and bills receivable ? 

e. What is each partner’s share of these book assets ? 

Corporations — Stocks and Bonds 

239. Men wishing to engage in a large business under¬ 
taking usually organize a Joint Stock Corporation, To do 
so, they must comply with the laws of the state in which 
they organize. A copy of the statutes relating to private 
corporations can be obtained by applying to the secretary 
of state in any state. 

240. The ownership of a corporation is divided into 
Shares of Stock. Stock is either Common Stock or Preferred 
Stock. The owner of each share of common stock is entitled 
to one vote in the management of the corporation and to 
dividends when and only when all other obligations of the 
company are paid. The owner of preferred stock is not 
entitled to any votes in the management of the business, 
unless it be the matter of issuing additional stock; he is 
entitled, however, to Dividends, usually at a specified rate 
per cent, before any dividends are paid on common stock. 

The preferred stock may be Cumulative Preferred, in which 
case the owner is guaranteed income at a fixed rate per cent 
per year; if the dividend for any year or years is unpaid 
when due, the owner is guaranteed that it will be paid 
ultimately if the corporation is solvent. 

Each share of stock has a nominal value, called the Par 
Value as a rule. Recently, corporations have been issuing 
common stock having no par value. Often but not always, 
the par value is $ 100. 

The shares of stock are sold in order to obtain the Capital 
of the corporation. 


340 


NEW HIGH SCHOOL ARITHMETIC 


Each purchaser of stock is given a Stock Certificate stat¬ 
ing the number and kind of shares purchased, and other 
information about them. 

The owner of shares of stock may sell them to some one 
else, by indorsing his stock certificate. The new purchaser 
then sends the indorsed certificate to the transfer agent 
of the corporation in order to have a record of the transfer 
made upon the books of the corporation. 

241. The Market Value of Stock depends upon various 
factors. Among these are the present net earnings of the 
corporation, the future prospects of it, and the current rate 
of interest paid upon sound investments. 

When a share of stock is purchased, the buyer acquires 
ownership of the next dividend payable on the share of 
stock, — unless the stock is sold ex dividend; in this latter 
event, the seller retains the rights to the dividend. 

Stocks that sell for more than their par value are said to 
be Above Par, and those that sell for less, Below Par. The 
price at which a share sells is called the Market Price. 


EXERCISE 146 

What is the par value and what is the market value of: 


Ex. No. 

No. op Shares 

Par Value 

Market Value 

1 . 

40 

|100 

1300 

2. 

15 

100 

101 

3. 

20 

100 

118 

4. 

50 

10 


5. 

25 

100 

58 

6. 

50 

No par value 

68J 

7. 

75 

15 

46i 

8. 

10 

100 

69| 

9. 

5 

100 

460 

10. 

20 

100 

162 










PARTNERSHIPS AND CORPORATIONS 341 


242 . The actual rate of income to the owner of a share of 
stock depends upon the price he paid for the stock and the 
present annual dividends received from it. To find the 
present rate of income on a share of stock, find what per cent 
the current annual dividends are of the current market 
price. Notice that the rate of income has nothing whatever 
to do with the par value. 

Example. What was the annual rate of income on a 
share of Anaconda Copper when the market price was 
$91.75 and the dividends were $ 8 per year? 

Solution. 1. What per cent of $ 91.75 is $ 8? 

2. $8 is 8.71% of $91.75. 


EXERCISE 147 

Eind the annual rate of income when one share of: 


Ex. No. 

Name of Company 

Was Sell¬ 
ing Foe 

And Annual 
Dividends 
Were 

1 . 

American Tobacco Common 

1172 

$20 

2 . 

Buckeye Pipe Line Common 

83 

14 

3 . 

Delaware and Hudson R. R. Common 

97^ 

9 

4 . 

Southern Pacific Common 

80 

6 

5 . 

Eureka Pipe Line Common 

245 

24 

6 . 

Federal Sugar Refining Pfd. 

95 

6 

7 . 

B. F. Goodrich Common 

72 

4 

8 . 

Norfolk and Western Pfd. 

89 

4 

9 . 

United Cigar Stores Common 

101 

7 

10 . 

Vacuum Oil 

355 

6 

11 . 

American Beet Sugar Common 

87 

8 

12 . 

Colorado Fuel and Iron Co. 

46 

3 

13 . 

New York Central Common 

80 

5 

14 . 

National Transit Co. 

14 

2 

15 . 

Savage Arms Co. 

55 

6 















342 


NEW HIGH SCHOOL ARITHMETIC 


243. Buying and Selling Stocks. Stocks are bought and 
sold through stock Brokers, located in large cities. They, 
in turn, or their clerks, meet at certain hours on business 
days at the Stock Exchange of their city. The procedure 
involved in buying and selling follows : 

Mr. Jones wants to sell 100 shares of U. S. Steel Common 
Stock. He gives his broker an order to that effect, telling him to 
sell “ at the market ” or at some specified price. If the market 
price happens to be 116^, Mr. Jones may ask 116£ for his stock. 
Mr. Smith may have decided to buy 75 shares, and Mr. Cummings 
25 shares of Steel Common, and ask their brokers to get it for 
them. At the stock exchange, the clerk of Mr. Jones’ broker 
offers 100 shares for sale, announcing the price; the clerks of the 
other brokers hid some other price. If a sale is arranged, Mr. 
Jones indorses his stock certificates over to the purchaser, who 
arranges for the transfer of the ownership on the books of the 
corporation. Mr. Jones’ certificate is canceled and new ones are 
made out for the purchasers and sent to them through their 
brokers. 

The brokers’ pay for their services is called brokerage. 
Shares valued from $1 to $10: brokerage is 7^ per share. 

Shares valued from $ 10 to $ 25 : brokerage is 12| f per share. 
Shares valued from $25 to $ 50: brokerage is 15^ per share. 
Shares valued from $ 50 to $75 : brokerage is 17| f per share. 
Shares valued from $ 75 to $ 100 : brokerage is 20 t per share. 
Shares valued from $ 100 to $200: brokerage is 25 $ per share. 

Minimum brokerage on the New York Exchange is $ 5. 

Example 1. What are the proceeds to the seller of 100 
shares of U. S. Steel Common Stock sold at $ 38.25 ? 

Solution. 1. Each share sells for $38.25. 

2. . *. 100 shares sell for 100 x $ 38.25, or $ 3825. 

3. The brokerage is 100 x $.15, or $ 15. 

4. .-. the proceeds = $3825 — $15, or $3810. 

Example 2. What is the cost to the buyer in Example 1 ? 

Solution. 1. The cost = $3825 -f $15, or $3840. 


PARTNERSHIPS AND CORPORATIONS 343 


EXERCISE 148 


Below are the quotations for certain stocks on the New 
York Stock Exchange on a certain day in 1931. 


Name of Stock 

Dividend in 
Dollars per 
Year 

High 

Price 

Low 

Price 

Clos¬ 

ing 

Price 

Allis Chalmers. 

1 

164 

I64 

I64 

American Locomotive, Pfd. . . 

7 

61| 

64 

64 

American Smelting, Pfd. . . . 

1.50 

24i 

23| 

234 

Baldwin Locomotive .... 

0 

8 | 

8 f 

84 

Bethlehem Steel, Pfd. 

7 

86 

85| 

854 

General Electric. 

1.60 

30£ 

28f 

*294 

Great Northern, Pfd. 

4 

27 

254 

254 

Norfolk and Western .... 

10 

1324 

130 

132i 

Pullman Co. 

3 

294 

274 

274 

Republic Iron and Steel . . . 

0 

7 

64 

6| 

Studebaker Auto Co. 

1.20 

12 

11 

11 

Union Carbide. 

2.60 

34f 

32| 

32| 

U. S. Steel. 

4 

684 

654 

65| 

Westinghouse Electric .... 

2.50 


44 

44 


Find the cost to the buyer and the proceeds to the seller 
in the sale of: 


Ex. No. 

No. OF 
Shares 

Name of Stock 

At Price 

1 . 

100 

American Locomotive Co. 

Low 

2 . 

50 

Bethlehem Steel, Pfd. 

High 

3 . 

20 

Allis Chalmers 

Close 

4 . 

40 

Great Northern, Pfd. 

High 

5 . 

75 

Republic Iron and Steel 

Low 

6 . 

200 

Studebaker 

High 

7 . 

75 

Union Carbide 

Low 

8 . 

250 

U. S. Steel 

Close 

9 . 

125 

Pullman Co. 

High 

10 . 

150 

Westinghouse Electric 

Low 





































344 


NEW HIGH SCHOOL ARITHMETIC 


Find the gain or loss by a stock speculator who 


Ex. No. 

Bought 

And Sold at 

No. of Shares 

Name of Stock 

At 

11. 

100 

U. S. Steel 

Low 

High 

12. 

200 

Westinghouse Electric 

Low 

High 

13. 

1500 

American Smelting Co. Pfd. 

Low 

High 

14. 

500 

Baldwin Locomotive 

High 

Low 

15. 

1000 

Great Northern Pfd. 

Low 

High 


16. A man having $ 2500 to invest decides to buy as 
many shares as he can of U. S. Steel Common. How many 
shares can he safely order through his broker? (Use the 
quoted “ high ” price.) 

17. A man having $ 3750 to invest decides to put i into 
Bethlehem Steel Pfd., 1 into American Smelting Co. Pfd., 
and i into General Electric. How many shares of each 
can he order ? 

18. How many shares of Norfolk and Western can a 
man buy who has $ 1500 to invest ? 

19. A man having $ 5000 to invest decides to place -f in 
Union Carbide, -J- in Studebaker Common, and f in 
Westinghouse Electric. How many shares of each can he 
order ? 

20. The trustees of an estate wish to invest $ 35,000, 
distributing the investment so that f is invested in pre¬ 
ferred stocks and -J in standard common stock. They 
decide to buy American Smelting Pfd., Great Northern Pfd., 
and Bethlehem Steel Pfd., investing approximately equal 
amounts in each. For common stock, they select Baldwin 
Locomotive, General Electric, and Republic Iron and Steel, 
investing about equal amounts in each. Find the number 
of shares of each stock that they can order. 












PARTNERSHIPS AND CORPORATIONS 345 


Bonds 

244. A Bond is a certificate indicating ownership of part 
of an indebtedness of a corporation; it is a promise to pay 
a fixed sum of money at a definite future time, with interest 
at a fixed rate, payable periodically. Often there is some 
sort of security pledged to secure payment of the bonds. 

Thus, the owners of a piece of property, wanting a large 
sum of money, borrow it from some banking institution, 
and, as security, mortgage their property to the institution 
as trustees. 

The bankers then issue bonds whose total face value is 
the amount of the sum loaned the owners of the property. 

On the other hand, governments of nations, states, and 
municipalities issue bonds, the only security for the pay¬ 
ment of which is the honor of the issuing government. 

The rate of interest and the time when the loan is repay¬ 
able are specified in the bond. The interest is generally 
payable semi-annually. It is a rate per cent of the face of 
the bond. 

A Registered Bond provides a place in which the owner’s 
name may be written. The borrower then mails the in¬ 
terest to the owner of the bond when it falls due. 

A Coupon Bond has coupons for each of the interest pay¬ 
ments due upon it attached to it. As these become due, 
the owner of the bond cuts them off and deposits them in 
his bank for collection. 

A complete description of the various kinds of bonds is 
beyond the scope of this book. 

Note. — In the section of a magazine, like World's Work , devoted 
to financial advertisements, investment houses often advertise their 
willingness to send without charge a copy of a pamphlet describing 
Stocks and Bonds. One such copy for a school is enough. 

Some banker in the town can probably give to the school a 
copy of a recent number of the Bond Record , published by Fitch 
Publishing Co., 138 Pearl St., New York City. 


346 


NEW HIGH SCHOOL ARITHMETIC 


245. Bonds are described in snch manner as to indicate 
snch items as tbe name of the debtor, the year of maturity, 
the rate of interest, and often the character of the security. 

Thus, Western Pacific R. R. Co. 1st Mortgage 5 J s, Series 
A, is a 5 °/o 1st mortgage bond of the Western Pacific 
R. R. Co. 

246. The great advantage of bonds, in general, over 
stocks is that they represent a definite sum of money that 
must be paid at a certain time; that they bear interest at a 
definite rate that must be paid at specified times, taking 
precedence over all dividend payments of the debtor. 

247. Cost of Bonds. The current market price of bonds 
may be above or below the face value of the bond, — above 
or below par. Whatever the market price, the owner of a 
bond is entitled to the face value of the bond at maturity. 

Bonds sell at a certain market price plus accrued interest. 

Thus, if a man buys on March 1 a bond on which the interest is 
payable on January 1 and July 1, he must pay the market price 
plus interest on the face of the bond from January 1 to March 1. 
On July 1 he receives the interest on the face from January 1 to 
July 1, thus getting the interest he paid the previous owner of 
the bond as well as the interest due on the bond since he pur¬ 
chased it. 

When bonds are purchased through a broker, Brokerage 
amounting to | % of the face of the bond is charged. 

Hence the purchase price equals market price plus ac¬ 
crued interest plus brokerage. 

248. The Rate of Income from a bond is the rate per 
cent which the annual interest is of the purchase price. 

Example. Find the purchase price of a $ 1000 5 % 
bond at 95 on March 12, if the interest is payable Feb. and 
Aug. 1. Also find the annual rate of income. 


PARTNERSHIPS AND CORPORATIONS 


347 


Solution. 1. The market price = 95 °fo of $ 1000 = $ 950. 

2. The accrued interest on $ 1000 from Feb. 1 to March 12 — 
that is, for 39 days — is $5.42. 

3* The brokerage is | °fo of $ 1000 or $ 1.25. 

4. Hence the purchase price = $ 950 + $ 5.42 + $ 1.25 = $ 956.67. 

5- The annual income is 5 °fo of $ 1000 or $ 50. 

6. The annual rate of income is 5.2+ #>, 

for $ 50 is 5.2+ % of $ 956.67. 

249. The Yield of a Bond is the rate of income realized 
from the bond if the bond is held until maturity. 

Thus, suppose a man buys a $ 100 bond for $ 98; that the rate 
of interest is 5 °fo and that the bond matures in 1 year. During 
the year, the man receives $5 interest (5 °Jo of $100), and at 
the end of the year $ 2 more than he paid for the bond, for the 
bond is worth $ 100 at maturity. All together he receives $ 7 on 
an investment of $98 for 1 year. His rate of income is about 
7.1 <f 0 . 

If this same bond matures in two years, he will have received a 
total of $ 10 interest and $ 2 profit when the bond matures. All to¬ 
gether he will have received $ 12 for the use of $ 98 for two years, 
or at the rate of $ 6 per year. This is about 6.1 °/o. This result is 
the approximate yield of income on this bond, as compound inter¬ 
est on the interest paid semi-annually has been neglected. 

Note. — The computation of yields on bonds is quite difficult. 
Investment houses have tables showing the rate yielded by bonds 
at various rates of interest for bonds maturing at the ends of 
various periods of years. 



































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TABLES OF MEASURES 

Measures of Length or Linear Measure 

12 inches (in.) = 1 foot (ft.) 

3 * eet = 1 yard (yd.) 

H yards or 16^ feet = 1 rod (rd.) 

320 rods ] 

1760 yards \ = 1 mile (mi.) 

5280 feet J 

1 knot or nautical mile = 6080 ft. (U. S.) 

Surveyors, in measuring land, use a chain (ch.) the length of 
which is 4 rods, divided into 100 links (li.) of 7.92 inches each. 
80 chains = 1 mile. 


Measures of Surface or Square Measure 


144 square inches (sq. in.) 

9 square feet 
60^ square yards 
160 square rods 1 
43560 square feet / 

640 acres 
10 square chains 
36 square miles 


— 1 square foot (sq. ft.) 
= 1 square yard (sq. yd.) 
= 1 square rod (sq. rd.) 

= 1 acre (A.) 

= 1 square mile (sq. mi.) 
= 1 acre 
= 1 township 


Measures of Volume or Cubic Measure 

1728 cubic inches (cu. in.) = 1 cubic foot (cu. ft.) 

27 cubic feet = 1 cubic yard (cu. yd.) 

128 cubic feet (a pile 4 ft. high, 4 ft. wide, and 8 ft. long) 
= 1 cord 


Liquid Measures 

4 gills (gi.) = 1 pint (pt.) 

2 pints = 1 quart (qt.) 

4 quarts = 1 gallon (gal.) 

It has been customary to consider 311 gallons the contents of a 
barrel (bbl.) and 63 gallons the contents of a hogshead. lu practice, 

349 


350 


NEW HIGH SCHOOL ARITHMETIC 


however, barrels are of various sizes; thus the standard gasoline 
barrel contains 56 gallons. A liquid quart contains 57f cubic inches; 
a liquid gallon 231 cubic inches. 


Dry Measures 

(Used in measuring grain, vegetables, fruits, etc.) 

2 pints (pt.) = 1 quart (qt.) 

8 quarts = 1 peck (pk.) 

4 pecks = 1 bushel (bu.) 

A dry quart contains 67£ cubic inches; a bushel contains 2150.42 
cubic inches. 


Avoirdupois Weight 


16 ounces (oz.) = 1 

100 pounds = 1 

20 hundredweight 1 _ ^ 
2000 pounds } — 


pound (lb.) 

hundredweight (cwt.) 

ton (T.) 


At United States Custom Offices, and at iron and coal mines, the 
following weights are used : 

112 pounds = 1 long hundredweight 
2240 pounds = 1 long ton 


Troy Weight 

(Used in weighing gold, silver, and jewels) 
24 grains (gr.) = 1 pennyweight (pwt.) 

20 pennyweights = 1 ounce (oz.) 

12 ounces = 1 pound (lb.) 

Apothecaries’ Weight 

20 grains (gr.) = 1 scruple (3) 

3 scruples = 1 dram (3) 

8 drams = 1 ounce (3 ) 

12 ounces = 1 pound (ft) 

Apothecaries’ Liquid Measures 

60 minims (m.) = 1 fluid dram (f 3) 

8 fluid drams = 1 fluid ounce (f 3 ) 

16 fluid ounces = 1 pint (O) 


TABLES OF MEASURES 


351 


Measures of Time 
60 seconds (sec.) = 1 minute (min.) 


60 minutes 

= 1 hour 

(hr.) 

24 hours 

= 1 day 

(da.) 

7 days 

= 1 week 

(wk.) 

52 weeks 1 


O r 0 

365 days J 

= 1 year 

366 days 

= 1 leap year 


If the number of a year is divisible by 4 but not by 100, the year is 
a leap year ; also when the number of the year is divisible by 400 the 
year is a leap year. In leap years, February has 29 days. 

The Solar Year has nearly 365| days. 

Angle and Arc Measures 
60 seconds (") = 1 minute (') 

60 minutes = 1 degree (°) 

A quadrant contains 90 arc degrees : a circle contains 360 arc 
degrees. A right angle contains 90 angular degrees. 

The length of a degree (jfo) of the earth’s equator is about 691 
miles. 

English Money 

4 farthings (far.) = 1 penny (d.) 

12 pence = 1 shilling (sh.) 

20 shillings = 1 pound or sovereign (£) 

= $4.8665 

French Money 

10 centimes (ct.) = 1 decime (dc.) 

10 decimes = 1 franc (fr.) 

1 franc = 19.3 cents 


German Money 
100 pfennige = 1 mark 

1 mark = 23.8 cents 


Miscellaneous 

Numbers 

12 units = 1 dozen (doz.) 

20 units = 1 score 
12 dozen = 1 gross (gr.) 

12 gross = 1 great gross 


Measures 

Paper 

24 sheets = 1 quire 
20 quires = 1 ream 
2 reams = 1 bundle 
5 bundles = 1 bale 


352 


NEW HIGH SCHOOL ARITHMETIC 


Carat. 


Some Standard Weights 

(Used in most of the States) 


Wheat 




. 60 lb. = 1 bushel 

Rye . 




. 56 lb. = 1 bushel 

Oats . 




. 32 lb. = 1 bushel 

Barley 




. 48 lb. = 1 bushel 

Corn, on cob 




. 70 lb. = 1 bushel 

Corn, shelled 




. 56 lb. = 1 bushel 

Potatoes . 




. 60 lb. = 1 bushel 

Water 

# 


. 

. 1 cu. ft. = 62| lb. 


(1) A term used to express the purity of gold. Pure gold =s 

24 carats fine. If 2, 4, etc., parts of alloy are present, 
the gold is 22, 20, etc., carats fine. Thus, 14 carat 
gold, used for good chains, contains 14 parts of gold to 
10 parts of alloy. 

(2) Also used as a unit for measuring precious stones. 

1 carat = 4 grains = about 3£ Troy grains. 

Since 1905, 1 carat = 200 milligrams, among inter¬ 
national jewelers. 








INDEX 

The numbers refer to pages 


Acceptance, 326 

Account purchases, 239; sales, 
239 

Acute angle, 146 
Addends, 4 

Addition, horizontal, 12 
Ad valorem duty, 295 
Aliquot parts, 78 
Altitude, of cylinder, 175; of 

frustum of cone, 195; of 
frustum of pyramid, 191; of 
parallelogram, 148; of par- 

allelopiped, 169; of prism, 
189; of rectangle, 148; of 

right circular cone, 194; of 

triangle, 148 
Amount, 245 

Angle, 145; acute, 146; obtuse, 
146; sides of, 146; vertex of, 
146 

Arc, 165 

Area, of circle, 166; of parallel¬ 
ogram, 158; of polygon, 148; 
of rectangle, 150; of solid, 168; 
of sphere, 178; of trapezoid, 
161; of triangle, 159, 164 
Assessors, 289 
Assets, 335 

Associative law of multiplication, 
25 

Bank, discount, 313; draft, 323 
Base, 134; of cone, 194; of 
parallelogram, 148; of par- 
allelopiped, 169; of pyramid, 


190; of rectangle, 148; of 
triangle, 148 

Bases, of cylinder, 175; of 
frustum of a cone, 195; of 
frustum of a pyramid, 191; 
of a prism, 189 

Base line in land surveying, 205 
Bill, 34; of exchange, 331 
Board foot, 173 
Bond, 345 

Brokerage, 239, 342, 346 
Brokers, 239 

Capital, 335 
Carpeting, 156 
Cashier’s check, 320 
Center of sphere, 178 
Certified check, 319 
Check, 317; cashier’s, 320; certi¬ 
fied, 319 
Circle, 165 
Circumference, 165 
Co-insurance, 280 
Collateral note, 306 
Commercial draft, 324 
Commission, 239; merchants, 
239 

Common, factor, 45; multiple, 
49; stock, 339 

Commutative law of multiplica¬ 
tion, 25 

Composite integer, 43 
Compound interest, 258; table 
of, 265 

Concrete work, 172 


353 





354 


INDEX 


Cone, 194 

Conical surface, 194 
Consignment, 239 
Corporation, 339 
Coupon bond, 345 
Credit, 303 

Cube, of a number, 134* root, 
135 

Cumulative preferred stock, 339 
Cylinder, 175 

Date of maturity of a note, 303 
Dates, time between, 115 
Days of grace, 304 
Decimal, fraction, 85; point, 85 
Decimals, adding, 88; subtract¬ 
ing, 88; multiplying, 90; di¬ 
viding, 95 

Degree of angle, 146 
Demand note, 303 
Denominate number, 108 
Denominate numbers, adding and 
subtracting, 113; multiplying, 
117; dividing, 118 
Denominator, 51 
Diagonal, of polygon, 147; of 
polyhedron, 189 

Diameter, of circle, 165; of 
sphere, 178 
Difference, 17 
Discount, 226 

Discounting, acceptances, 329; 
notes, 313 

Distributive law of multiplica¬ 
tion, 25 
Dividend, 37 

Dividends, in insurance, 286; 

on stock, 339 
Divisor, 37 
Draft, 322 

Drawee, of check, 317; of draft, 
323 

Drawer, of check, 317; of draft, 
323 

Duties, 295 


Edges of polyhedron, defined, 
189 

Even number, 43 
Evolution, 135 
Exchange, 317 
Exponent, 134 

Face, of check, 317; of note, 303; 

of polyhedron, 189 
Factor, 43 

Factoring integers, 43 
Figures, geometrical, 145 
Finding, principal in interest, 
256; rate in interest, 253; 
time in interest, 255 
Foreign money, 130 
Formula, 142, 150 
Fraction, common, 51; improper, 
51; proper, 51 
Fractional parts, 75 
Fractions, adding, 60; subtract¬ 
ing, 63; multiplying, 65; di¬ 
viding, 71; decimals, 85; fun¬ 
damental laws of, 52; lowest 
terms of, 54 
Franc, 130 

Frustum, of cone, 195; of pyra¬ 
mid, 191 

Geometrical figures, 145 
Gram, 125 
Graph, 198 

Greatest common divisor, 45 
Gross, cost, 234,239; proceeds, 239 

Hypotenuse, 162 

Improper fraction, 51 
Income taxes, 292, 298 
Index of radical, 135 
Inheritance taxes, 294, 301 
Installment payments, 269 
Insurance, 272; automobile, 282 ; 
fife, 284; premium, 272; 
policy, 272 



INDEX 


355 


Integer, 1; composite, 43; prime, 
43 

Integers, adding, 4; subtracting, 
17; multiplying, 24; dividing, 
37; factoring, 43 
Interest, 245; compound, 258; 
periodic, 267; simple, 245; 
computing, 245; days, 259; 
period, 259 

Interest-bearing note, 304 
Invoice, 34, 227 
Involution, 134 

Joint note, 305 

Kilogram, 126 
Kiloliter, 125 
Kilometer, 123 

Lateral area, of cylinder, 175, 
176; of frustum of a cone, 
195; of frustum of a pyramid, 
192; of prism, 189; of pyra¬ 
mid, 190, 191; of right circular 
cone, 195 

Lateral edges, of a prism, 189; 

of a pyramid, 190 
Lateral faces, of a prism, 189; 

of a pyramid, 190 
Lateral surface of a right circular 
cone, 194 

Least common multiple, 49 
Liabilities, 335 

Life insurance, 284; dividends, 
286 

Line, broken, 145; closed, 145; 

curved, 145; straight, 145 
Lines, parallel, 146 
List price, 226 
Liter, 125 

Maker of a note, 303 
Making change, 22 
Mark, 130 
Marking goods, 237 


Matched lumber, 182 
Measure, 108 
Measuring lumber, 173 
Merchants’ rule of partial pay¬ 
ments, 308 
Meter, 123 
Metric measures, 123 
Minuend, 17 
Mixed number, 51 
Multiple, 49 
Multiplicand, 24 
Multiplier, 24 t 

Negotiable note, 305 
Net cost, 234 
Net proceeds, 239 
Non-negotiable note, 305 
Number, denominate, 108; 
mixed, 51 

Numbers, Roman system of, 3 
Numerator, 51 

Obtuse angle, 146 

Papering, 154 
Parallel lines, 146 
Parallelogram, 148 
Parentheses, 24. 

Partial payments, 307 
Partner, general, 335 
Partnership, 335 
Par value of stock, 339 
Payee, of check, 317; of draft, 
323; of note, 303 
Per cent, 208 
Percentage, 208 
Perimeter of polygon, 147 
Periodic interest, 267 
Periods in square root, 137 
Perpendicular, 146 
Plastering, 153 
Polygon, 147; area of, 148 
Polyhedron, 189 
Pound (English money), 130 
Power of a number, 134 



356 


INDEX 


Powers of ten, 85 
Preferred stock, 339 
Prime, cost, 239; factors, 43; 
integer, 43 

Principal, 239, 245; meridian, 
205 

Prism, 189 

Proceeds of discounted note, 313 
Product, 24 
Profits and losses, 234 
Promissory note, 303 
Proper fraction, 51 
Property tax, 287 
Protesting a note, 305 
Pyramid, 190 

Quadrilateral, 148 
Quotient, 37 

Radical sign, 135 
Radius, of circle, 165; of sphere, 
178 

Range, 206 

Rate, of discount, 313; of do¬ 
mestic exchange, 329; of in¬ 
come on bonds, 346; of 
interest, 245; per cent, 208 
Ray, 145 
Rectangle, 148 

Rectangular parallelopiped, 169 
Registered bond, 345 
Regular pyramid, 190 
Remainder in division, 38 
Right angle, 146; prism, 189; 
triangle, 162 

Roman system of numbers, 3 
Roofs and roofing, 184 

Savings bank, 259 
Segment of line, 145 
Sequence of integers, 4 
Several note, 305 
Share of stock, 339 
ShiUing, 130 
Shipment, 239 


Short methods of multiplication, 
29 

Sides, of an angle, 146; of a 
polygon, 147 
Sight draft, 324 
Simple interest, 245 
Six per cent method, 247 
Slant height, of frustum of a 
cone, 195; of regular pyramid, 
191; of right oircular cone, 194 
Solid, 168 
Specific duty, 295 
Sphere, 178 

Square, 148; of a number, 134; 
root, 135, 136 

Standard fire insurance policy, 
273 

Stock certificate, 340 
Subtrahend, 17 
Successive discounts, 227 
Sum of integers, 4 
Surface, closed, 168; conical, 
194; spherical, 168 

Tables of powers and roots, 140 
Tables of measures, 349 
Tables, compound interest, 265 
Tax, real estate, 287 
Term of discount, 313 
Terms of sale, 227 
Time, between dates, 115; draft, 
325; note, 304 
Township, 206 
Trapezoid, 160 
Triangle, 148 

Unit on a graph, 198 
United States rule of partial pay¬ 
ments, 310 

Units of measure, 108 

Vertex, of angle, 146; of pyra¬ 
mid, 190; of right circular 
cone, 194 



INDEX 


357 


Vertices, of polygon, 147; of 
polyhedron, 189 

Volume, of a cone, 195; of 
f rust rum of a cone, 195; of 
frustrum of a pyramid, 192; of 
parallelopiped, 169; of prism, 


190; of pyramid, 192; of right 
circular cylinder, 176; of solid, 
168; of sphere, 178 

Yield of a bond, 347 


























. 
















(t 



* 


























